Hi Brilliant! This is the sequel of the first part of Brilliant Summation Contest Season1.
Update: This contest has been ended. Thanks for everyone's participation.
The aims of the Summation contest are to improve skills in the computation of sums, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.
The rules are as follows:
I will start by posting the first problem. If there is a user solves it, then they must post a new one.
You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
Only make substantial comment that will contribute to the discussion.
Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.
If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
You are NOT allowed to post a multiple summation problem.
Problems must be purely of summation. They shouldn't have integrals and products in them. Solutions can follow methods that use integrals and products.
It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.
There is no restriction in the standard of summations.
Please post your solution and your proposed problem in a single new thread.
Format your post is as follows:
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Top NewestProblem 27:
Prove:
\[\sum _{ n=\infty }^{ \infty }{ { \left( e^{\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \frac { \pi }{ 2 } \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \] – Aditya Kumar · 1 year ago
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\[\sum _{ n=\infty }^{ \infty }{ { \left( e^{\pi} \right) }^{ { n }^{ 2 } } } =\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \]
Now use this formula \[\phi(e^{2\pi/n})=\frac { { \left( \pi \right) }^{ \frac { 1 }{ 4 } } }{ \Gamma \left( \frac { 3 }{ 4 } \right) } \frac{n^{1/4}r_{2,2/n^2}}{2^{1/4}r_{2,2}}\]
\[r_{k,n} \] is a function of DedeKind Eta function. Which have some formulaes to solve.
\[r_{2,2}=2^{1/8} \text{and} r_{2,1/2}=2^{1/8}\]
Thus , you get \[\frac{(\frac{\pi}{2})^{1/4}}{\Gamma(3/4)}\] – Aman Rajput · 1 year ago
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– Aditya Narayan Sharma · 1 year ago
This is evaluated in ramanujan's notebook which I have but there's no proof for the formula used :/ . So I was trying to prove the above qproductLog in to reply
– Aman Rajput · 1 year ago
share your q product .... if i have time i can solve it easilyLog in to reply
– Aditya Narayan Sharma · 1 year ago
No, I m not talking bout the qproduct actually. I m talking about the above formula , he used qproducts and generalisations to come to this. So share if you have a proof of this , coz I didnt findLog in to reply
– Aditya Kumar · 1 year ago
At least could you elaborate on step 2?Log in to reply
– Aman Rajput · 1 year ago
I used this identity from a ramanujan paperLog in to reply
Problem 18:
Prove that:
\[\sum_{k=0}^n \binom{n+k}{nk}\frac{(1)^{n+k}}{2k+1}=\frac{2}{2n+1}\,\cos\left(\frac{2(n1)\pi}{3}\right)\;\text{?}\]
This Problem has been solved by Ishan Singh. – Aditya Kumar · 1 year ago
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Proof : Since \( \dbinom{n}{r} = 0 \) for \( r > n \), we can rewrite the sum as
\[ \text{S}_{n} = \sum_{r=0}^{\infty} (1)^r \dbinom{2n+1r}{r} \]
From the Binomial Theorem, we see that the sum is the coefficient of \(x^n\) in
\[f(x) = x^n (1x)^{2n+1} + x^{n1} (1x)^{2n} + \ldots \]
\[ = \dfrac{(1x)^{n+1}}{x^2x+1} \]
Note that,
\[ \sum_{n=0}^{\infty}{{U}_{r}(a) {x}^{r}} = \dfrac{1}{x^2 2ax+1} \]
where \( U_{r} (x) \) is the Chebyshev Polynomial of the second kind.
Putting \(a = \dfrac{1}{2}\), we see that,
\[ f(x) = (1x)^{n+1} \sum_{n=0}^{\infty}{{U}_{r} \left( \dfrac{1}{2} \right) {x}^{r}} \quad (*) \]
To calculate the coefficient of \(x^n\) in \( (*) \), we see that coefficient of \(x^{nr}\) for fixed \(r\) in \((1x)^{n+1}\) is \( (1)^{nr} \dbinom{n+1}{nr} \), so coefficient of \(x^n\) is,
\[ \sum_{r=0}^{n} (1)^{nr} \dbinom{n+1}{nr} U_{r} \left(\dfrac{1}{2}\right) \]
Substituting \( r \mapsto r1 \), we have,
\[ \text{S}_{n} = \sum_{r=1}^{n+1} (1)^{n+1r} \dbinom{n+1}{n+1r} U_{r1} \left(\dfrac{1}{2}\right) \]
\[ = (1)^{n+1} \sum_{r=1}^{n+1} (1)^{r} \dbinom{n+1}{r} U_{r1} \left(\dfrac{1}{2}\right) \quad \quad \left( \because \dbinom{n}{nr} = \dbinom{n}{r} \right) \]
\[ = (1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=1}^{n+1} (1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because U_{r1} \left(\dfrac{1}{2}\right) = \sin \left( \dfrac{r \pi}{3} \right) \right) \]
\[ (1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=0}^{n+1} (1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because \sin 0 = 0 \right) \]
Now, \[\displaystyle \sum_{r=0}^{n+1} (1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \]
\[ \implies \text{S}_{n} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square \]
Proof : \[\sum_{k=0}^n(1)^{n+k} \dbinom{n+1+k}{2k+1} = \sum_{k=0}^n(1)^{n+k} \dbinom{n+1+k}{nk} = \sum_{k=0}^n(1)^k \dbinom{2n+1k}{k} = \text{S}_{n} \]
\[\left( \because \dbinom{n}{r} = \dbinom{n}{nr} \ \text{&} \ \sum_{r=0}^{n} f(k) = \sum_{r=0}^{n} f(nk) \right) \quad \square \]
Now,
\[ \lambda = \sum_{k=0}^n\dfrac{(1)^{n+k}}{2k+1} \dbinom{n+k}{nk} \]
\[ = \sum_{k=0}^n \dfrac{(1)^{n+k}(n+k)!}{(2k+1)!(nk)!} \]
\[ =\dfrac{1}{2n+1}\sum_{k=0}^n\dfrac{(1)^{n+k}(n+k)!\left[(n+k+1)+(nk)\right]}{(2k+1)!(nk)!}\]
\[ =\dfrac{1}{2n+1}\sum_{k=0}^n\left((1)^{n+k} \dbinom{n+1+k}{2k+1}  (1)^{n1+k} \dbinom{n+k}{2k+1} \right) \]
\[ =\dfrac{1}{2n+1} \left(\text{S}_{n}  \text{S}_{n1}\right) \]
Using the Proposition and simplifying, we have,
\[ \lambda = \dfrac{2}{2n+1} \, \cos\left(\frac{2(n1)\pi}{3}\right) \quad \square \]
– Ishan Singh · 1 year ago
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Problem 17 :
Prove That
\[ \displaystyle \sum_{n=0}^{\infty} \dfrac{2^{n1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}} \]
This problem has been solved by Aditya Kumar. – Ishan Singh · 1 year ago
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\[S=\sum _{ n=0 }^{ \infty } \frac { 2^{ n1 }n! }{ \prod _{ r=0 }^{ n } (4r+1) } \]
\[S=\sum _{ n=0 }^{ \infty } \left[ \frac { { 2 }^{ n1 }\Gamma \left( n+1 \right) \Gamma \left( \frac { 5 }{ 4 } \right) }{ \Gamma \left( n+\frac { 5 }{ 4 } \right) } \right] \]
Using beta function, we get:
\[S=\sum _{ n=0 }^{ \infty } \left[ { 2 }^{ n1 }B\left( n+1,\frac { 5 }{ 4 } \right) .\left( n+\frac { 5 }{ 4 } \right) \right] \]
\[S=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ n }B\left( n+1,\frac { 5 }{ 4 } \right) .n \right] +\frac { 5 }{ 8 } \sum _{ n=0 }^{ \infty } \left[ { 2 }^{ n }B\left( n+1,\frac { 5 }{ 4 } \right) \right] \]
\[S=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ n }.n.{ t }^{ n }{ \left( 1t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} } +\frac { 5 }{ 8 } \int _{ 0 }^{ 1 }{ \left\{ \sum _{ n=0 }^{ \infty }{ \left[ { 2 }^{ n }.{ t }^{ n }{ \left( 1t \right) }^{ \frac { 1 }{ 4 } } \right] } dt \right\} } \]
Now I'll use the following geometric progression summations:
\[A=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ \left( 1t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { \left( 1t \right) }^{ \frac { 1 }{ 4 } } }{ 1xt } ,\quad B=\sum _{ n=0 }^{ \infty }{ \left[ { \left( xt \right) }^{ n }{ .n.\left( 1t \right) }^{ \frac { 1 }{ 4 } } \right] } =\frac { { xt\left( 1t \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 1xt \right) }^{ 2 } } \]
\[S=\frac { 5 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { { \left( 1t \right) }^{ \frac { 1 }{ 4 } } }{ 2t } dt } +\int _{ 0 }^{ 1 }{ \frac { x{ \left( 1x \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 2x \right) }^{ 2 } } dx } \]
The first integral can be easily evaluated. The second one is also easily evaluated. I was screwed up and I couldn't evaluate it. So I asked for it here.
Hence the final answer on doing some simplification is:
\[\boxed{\displaystyle\sum_{n=0}^{\infty} \dfrac{2^{n1} n!}{\prod_{r=0}^{n} (4r+1)} = \dfrac{\pi + 2\log(1+\sqrt{2})}{4 \sqrt{2}}}\] – Aditya Kumar · 1 year ago
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Problem 24:
Prove that \[\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}\]
This problem has been solved by Aditya Kumar. – Aman Rajput · 1 year ago
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Proof:
Consider: \({ f }_{ n }\left( x \right) =\frac { \sin ^{ 2n }{ x } }{ \left( 2n \right) ! } \).
Then we get: \(\\ { f" }_{ n }\left( x \right) ={ f }_{ n1 }\left( x \right) { \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) \)
We can write \(\displaystyle { x }^{ 2 }=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ f }_{ n }\left( x \right) } \) for some function \(a_n\).
On differentiating both sides twice:
\[2=\sum _{ n=1 }^{ \infty }{ { a }_{ n }\left( { f }_{ n1 }\left( x \right) { \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) \right) } =\sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ f }_{ n }\left( x \right) } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) } \]
From this relation \(a_1=2\) and \(a_{n+1}=a_n{(2n)}^{2}\)
Thus, \(\displaystyle \frac { { a }_{ n } }{ 2 } =\prod _{ k=1 }^{ n1 }{ \frac { { a }_{ k+1 } }{ { a }_{ k } } } ={ 2 }^{ 2n2 }{ \left( n1 \right) ! }^{ 2 }\).
Hence, we get : \(\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] } \)
Proof:
From Lemma 1, we can write: \(\displaystyle { x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n } }{ { n }^{ 2 } \dbinom{2n}{n} } \sin ^{ 2n }{ x } \right] }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) } \)
Here, \(b_n={ 2 }^{ n }(n1)!\) and \(b_{n+1}=2nb_n\).
We can write: \(\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ f }_{ n }\left( x \right) } \) for some function \(a_n\).
On differentiating both sides twice:
\[12{ x }^{ 2 }=\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ { a }_{ n+1 }{ b }_{ n+1 }^{ 2 }{ f }_{ n }\left( x \right) } \frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ { a }_{ n }{ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }{ f }_{ n }\left( x \right) } \]
Thus, we get: \(({ a }_{ n+1 }{ a }_{ n }){ b }_{ n }^{ 2 }{ \left( 2n \right) }^{ 2 }=12{ b }_{ n }^{ 2 }\)
From here we get: \(({ a }_{ n+1 }{ a }_{ n })=\frac { 12 }{ { \left( 2n \right) }^{ 2 } } \).
Hence, \({ a }_{ n }={ H }_{ n1 }^{ (2) }\).
Therefore, we get: \(\displaystyle { x }^{ 4 }=\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \left[ \frac { { 2 }^{ 2n }\sin ^{ 2n }{ x } { H }_{ n1 }^{ (2) } }{ { n }^{ 2 }\dbinom{2n}{n} } \right] }\).
From lemma 2, we get: \[\displaystyle (\sin^{1} x)^4 = \frac32 \sum_{n=1}^\infty \frac{2^{2n} H_{n1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2n} \ \quad ...(A)\]
Now, in the problem, we use: \({ F }_{ 2k }=\frac { { \varphi }^{ 2k }\frac { 1 }{ { \varphi }^{ 2k } } }{ \sqrt { 5 } } \)
On substituting and using \(A\), we get: \[\boxed{\displaystyle \sum_{k \geq 1} \frac{F_{2k}H_{k1}^{(2)}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt5}}\] – Aditya Kumar · 1 year ago
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– Ishan Singh · 1 year ago
(+1) Nicely explained! I solved using Beta Functions and Integration.Log in to reply
Problem 22:
Prove the identity:
\[\sum _{ k=1 }^{ n }{ \frac { { H }_{ k } }{ nk+1 } } ={ \left( { H }_{ n+1 } \right) }^{ 2 }{ H }_{ n+1 }^{ \left( 2 \right) }\]
This problem has been solved by Aditya Sharma, Mark Hennings and Ishan Singh. – Aditya Kumar · 1 year ago
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\[ \text{S} = \sum_{k=1}^{n}\dfrac{H_{k}}{nk+1} \]
\[ = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{r(nk+1)} \]
\[ = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r(nk+1)} \]
\[ = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{r(kr+1)} \]
\[ = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{(kr+1 + r)}{r(kr+1)(k+1)} \]
\[ = \sum_{r=1}^{n} \sum_{k=r}^{n} \dfrac{1}{(k+1)} \left[\dfrac{1}{(kr+1)} + \dfrac{1}{r}\right] \]
\[ = \sum_{k=1}^{n} \sum_{r=1}^{k} \dfrac{1}{(k+1)} \left[\dfrac{1}{(kr+1)} + \dfrac{1}{r}\right] \]
\[ = 2\sum_{k=1}^{n} \dfrac{H_{k}}{k+1} \]
Substitute \( k+1 \mapsto k \)
\[\implies \text{S} = 2\sum_{k=2}^{n+1} \dfrac{H_{k1}}{k} \]
\[ = 2\sum_{k=2}^{n+1} \dfrac{H_{k1}  \dfrac{1}{k}}{k} \]
\[ = 2\sum_{k=1}^{n+1} \dfrac{H_{k}}{k}  2H_{n+1}^{(2)} \]
Since \(\displaystyle \sum_{k=1}^{n+1} \dfrac{H_{k}}{k} = \dfrac{1}{2} (H_{n+1}^2 + H_{n+1}^{(2)}) \) (I have proved it here), we have,
\[\text{S} = (H_{n+1})^2  H_{n+1}^{(2)} \quad \square \] – Ishan Singh · 1 year ago
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We have , \(\displaystyle S(n+1) = \sum_{k=1}^{n} \frac{H_k}{nk+1} = \sum_{k=1}^{n}\frac{H_{nk+1}}{k} = \sum_{k=1}^{n}\frac{H_{nk}}{k} + \frac{2H_n}{n+1} = S(n)+\frac{2H_n}{n+1}\)
Telescoping above we have , \(\displaystyle S(n+1) = \sum_{k=0}^{n}\frac{2H_k}{k+1}\)
\(\displaystyle \text{Lemma :} \large \boxed{\sum_{k=0}^{n} [(k+1)^pk^p]H_{k}^{(m)} = (n+1)^p H_{n}^{(m)}H_{n}^{(mp)}}\)
\(\text{Proof :}\) Using summation by parts we have ,
\(\displaystyle \sum_{k=0}^{n} [(k+1)^pk^p]H_{k}^{(m)} = (n+1)^p H_{n+1}^{(p)}  \sum_{k=0}^{n} (k+1)^{(mp)} = (n+1)^p H_{n+1}^{(p)}  H_{n+1}^{(mp)} = (n+1)H_{n}^{(p)}  H_{n}^{(mp)}\) & thus proved.
We'll use a well known Harmonic sum identity , \(\displaystyle \sum_{k=1}^{n} \frac{H_k}{k} = \frac{1}{2}((H_n)^2+H_{n}^{(2)})\)
Applying the lemma for \(p=1,m=1\) we have,
\(\displaystyle S(n+1) = 2\sum_{k=0}^{n} \frac{H_k}{k+1} = \sum_{k=1}^{n} \frac{2H_k}{k} + \frac{H_n}{n+1}H_{n}^{(2)}\)
Simplifying we get , \(\displaystyle \sum_{k=1}^{n} \frac{H_k}{nk+1} = (H_{n+1})^2H_{n+1}^{(2)}\) – Aditya Narayan Sharma · 1 year ago
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– Mark Hennings · 1 year ago
The generating function of the \(S(n)\) is \[ \sum_{n=1}^\infty S(n) x^{n+1} = \sum_{n=1}^\infty H_n x^n \times \sum_{n=1}^\infty \tfrac{1}{n} x^n = \frac{\ln^2(1x)}{1x} \] so the result follows using the known formulae for the generating functions for \(H_n^{(2)}\) and for \((H_n)^2\).Log in to reply
Problem 29:
Prove:
\[\displaystyle \sum\limits_{m=1}^{99}{\frac{\sin{\left(\frac{17 m \pi}{100}\right)} \sin{\left(\frac{39 m \pi}{100}\right)}}{1+\cos{\left( \frac{m\pi}{100} \right) }}}=1037 \]
Due to time constraint, I have decided to post the solution. – Jack Lam · 1 year ago
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Observing that \(\cos{(k+1)\theta} + 2\cos{k\theta} + \cos{(k1)\theta} = 2\cos{k\theta}(1+\cos{\theta}) \)
The following recurrence relation is obtained:
\(\displaystyle T(k+1) + 2T(k) + T(k1) = 2 \sum_{m=1}^{n1} \cos{\frac{mk\pi}{n}} = 1+\cos{k\pi} + \sin{k\pi} \tan{\frac{k\pi}{2n}}\)
Note that an implicit restriction of \( k \notin n \mathbb{Z} \) is forced because the GP formula is invalid when the common ratio is \(\pm 1\).
Evaluate the recurrence relation at \(k+1\) and add together to yield
\( T(k+2) + 3T(k+1) + 3T(k) + T(k1) = 2 \)
Evaluate this recurrence relation at \(k\) and take the difference to homogenise the recurrence relation:
\( T(k+2) + 2T(k+1)  2T(k1)  T(k2) = 0 \)
The characteristic roots are \(1\) and \(1\) with multiplicity \(3\).
The general solution to the recurrence relation is thus
\( (C_1 k^2 + C_2 k+C_3)(1)^k + C_4\)
Returning to the original problem, define \(\displaystyle S(a,b,n) = \sum_{m=1}^{n1} \frac{\sin{\frac{am\pi}{n}}\sin{\frac{bm\pi}{n}}}{1+\cos{\frac{m\pi}{n}}}\) where \( 0 \le b \le a \le n \)
We find that \(2S(a,b,n) = T(ab)T(a+b)\)
Note that \( (1)^{a+b} \equiv (1)^{ab}\)
Substituting \(T(k)\) into the above and expanding, we obtain
\(\displaystyle S(a,b,n) = (2C_1 a + C_2)b(1)^{a+b+1} = (1)^{a+b}\hat{C_1}b(\hat{C_2}a)\)
A short computation yields \(S(1,1,n) = n1\)
From \( S(n,b,n) = 0 \, \& \, S(1,1,n)>0\), \(\hat{C_2} = n\)
From \(S(1,1,n) = n1\), \(\hat{C_1} = 1\)
Therefore, we conclude \( S(a,b,n) = (1)^{a+b} b(na) \)
Substituting \(b=17, a=39, n=100\), and the result which was to be shown follows. – Jack Lam · 1 year ago
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– Deeparaj Bhat · 1 year ago
This was the best solution!Log in to reply
– Aditya Kumar · 1 year ago
Nice solution. I missed the 48 hour time.Log in to reply
Problem 20 :
Evaluate \[\sum_{i_m=1}^n \sum_{i_{m1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2} 1 \]
This problem has been solved by Aditya Sharma and Ameya Daigavane. – Deeparaj Bhat · 1 year ago
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@Aditya Sharma has solved the question  I thought of an approach that might be considered more "illuminating" than induction.
AlthoughLet's think of the \( i_1, i_2,...i_m \) as numbers on a register  each \( i_k\) is one number from \(1\) to \(n\). Initially, the register reads \( m \) ones, because all the \(i_k\) are \(1\).
For each unique number on the register, we add one to our count.
Now, we turn the first number (\(i_m\)) to \(2\).
The register reads \(2\) followed by \(m  1\) \(1\)s.
Note that a digit to the right of another digit can never be greater than it, because \(i_j \geq i_k\)if \(j \geq k\).
So, the number on our register is an \(m\) digit number whose digits are nondecreasing when read from left to right.
We can keep turning the register and counting how many unique numbers we get, until we reach \(m\) digits with value \(n\). After this, our register stops working.
It is easy to see that every single number satisfying the nondecreasing digit condition is encountered, and only once.
So our question now becomes:
What is the number of tuples \( i_1, i_2,...i_m \) satisfying,
\[ 1 \leq i_1 \leq i_2 \leq \ldots \leq i_m \leq n \]
Because for each one of these tuples, we have a number on our register.
If we write the numbers in a row, \[ i_1 i_2 \ldots i_m \] and divide them into \(n\) blocks, such that the numbers enclosed in the \(n^{th}\) block get value \( n \), then we have found one such possibility. We need \(n  1\) bars to separate the numbers into \(n\) blocks.
For example, if \(m = 4, n = 4\), we can divide the numbers as, \[ i_1   i_2 i_3  i_4 \] Here 0 \[ i_1 = 1, i_2 = i_3 = 3, i_4 = 4 \]
Clearly, the number of ways we can separate these numbers, is the number of tuples we need, because numbers with higher indices are always given higher (or equal) values.
So, we can finish with starsandbars, as we have \(m\) numbers, and \(n 1 \) separators, so the total number of separations/tuples is: \[ \binom {m + n  1}{n  1} \] which is the same answer as above. – Ameya Daigavane · 1 year ago
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– Aditya Kumar · 1 year ago
Ok sorry for that. It is just that brilliant is the only website that doesn't use $. So that made me think that it was from other source.Log in to reply
– Ameya Daigavane · 1 year ago
I work with LaTeX on other sites as well. I had thought about this question for some time; in fact, I think there was a similar question in the FIITJEE GMP for JEE Advanced. I really liked my solution there, so I thought I'd just extend it here.Log in to reply
\(\displaystyle S=\sum_{i_m=1}^n \sum_{i_{m1}=1}^{i_m} \cdots \sum_{i_2=1}^{i_3} \binom{i_2}{1} \)
\(\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m1}=1}^{i_m} \cdots \sum_{i_3=1}^{i_4} \binom{i_3+1}{2}\)
\(\displaystyle S = \sum_{i_m=1}^n \sum_{i_{m1}=1}^{i_m} \cdots \sum_{i_4=1}^{i_5} \binom{i_4+2}{3} \)
It's obvious from induction that \(\displaystyle S=\sum_{i_m=1}^n \binom{i_m+m2}{m1} = \frac{n}{m}\binom{m+n1}{n}\) – Aditya Narayan Sharma · 1 year ago
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Problem 26: Prove : \[\displaystyle \sum_{k \geq 1}\frac{H_k^{(4)}H_{k\frac12}^{(4)}}{2k+1}=2\zeta(2)\zeta(3)(15\ln 214)\zeta(4)\]
This problem has been solved by Aditya Kumar. – Aman Rajput · 1 year ago
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\[\sum _{ n=1 }^{ \infty }{ \left[ \frac { { H }_{ n }^{ (4) }{ H }_{ \frac { 2n1 }{ 2 } }^{ (m) } }{ 2n+1 } \right] } =\sum _{ n=1 }^{ \infty }{ \left[ \frac { 16\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( 2n+2k1 \right) }^{ 4 } } } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( n+k \right) }^{ 4 } } } }{ 2n+1 } \right] } \]
\[=\sum _{ n=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \left[ \frac { 16{ \left( n+k \right) }^{ 4 }{ \left( 2n+2k1 \right) }^{ 4 } }{ \left( 2n+1 \right) { \left( n+k \right) }^{ 4 }{ \left( 2n+2k1 \right) }^{ 4 } } \right] } } \]
\[=\sum _{ n=1 }^{ \infty }{ \sum _{ k=n+1 }^{ \infty }{ \left[ \frac { 16{ \left( k \right) }^{ 4 }{ \left( 2k1 \right) }^{ 4 } }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k1 \right) }^{ 4 } } \right] } } \]
\[=\sum _{ n=1 }^{ \infty }{ \sum _{ k=n+1 }^{ \infty }{ \left[ \frac { 32{ \left( k \right) }^{ 3 }{ 24\left( k \right) }^{ 2 }+8k1 }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k1 \right) }^{ 4 } } \right] } } \]
\[=\sum _{ n=1 }^{ \infty }{ \left[ \sum _{ k=n }^{ \infty }{ \left[ \frac { 32{ \left( k \right) }^{ 3 }{ 24\left( k \right) }^{ 2 }+8k1 }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k1 \right) }^{ 4 } } \right] } \frac { 32{ \left( n \right) }^{ 3 }{ 24\left( n \right) }^{ 2 }+8n1 }{ \left( 2n+1 \right) { \left( n \right) }^{ 4 }{ \left( 2n1 \right) }^{ 4 } } \right] } \]
Now, we change the order of the double sum:
\[=\sum _{ k=1 }^{ \infty }{ \sum _{ n=1 }^{ k }{ \left[ \frac { 32{ \left( k \right) }^{ 3 }{ 24\left( k \right) }^{ 2 }+8k1 }{ \left( 2n+1 \right) { \left( k \right) }^{ 4 }{ \left( 2k1 \right) }^{ 4 } } \right] } } \sum _{ n=1 }^{ \infty }{ \left[ \frac { 32{ \left( n \right) }^{ 3 }{ 24\left( n \right) }^{ 2 }+8n1 }{ \left( 2n+1 \right) { \left( n \right) }^{ 4 }{ \left( 2n1 \right) }^{ 4 } } \right] } \]
Now using the 3 formulas I have provided above and evaluating each summation (which is basic wrt this contest), on simplification gives:
\[\displaystyle \sum_{k \geq 1}\frac{H_k^{(4)}H_{k\frac12}^{(4)}}{2k+1}=2\zeta(2)\zeta(3)(15\ln 214)\zeta(4)\]
The simplification is left as an exercise to the readers. – Aditya Kumar · 1 year ago
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@Aditya Kumar – Aman Rajput · 1 year ago
justify your last double summation. partial fraction is justified.Log in to reply
– Aditya Kumar · 1 year ago
See formula 2 for that.Log in to reply
Problem 25:
Evaluate:
\[\sum _{ r=1 }^{ n }{ { H }_{ r }^{ 2 } } \]
This problem has been solved by Aman Rajput. – Aditya Kumar · 1 year ago
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\[=\displaystyle (n+1)H_n^2nH_nnH_{n1}+n1+nH_n\] – Aman Rajput · 1 year ago
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\(\text{Problem 23 :}\)
Prove that : \(\displaystyle \sum_{n=\infty}^{\infty} \frac{(1)^n}{(n+a)^4} = \frac{\pi^4}{6} \frac{\cos(\pi a)}{\sin^2 (\pi a)} (6\csc^2 (\pi a) 1) \) for \(a \in \mathbb{R} \) and \(a\) is not an integer.
This problem has been solved by Aman Rajput. – Aditya Narayan Sharma · 1 year ago
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Consider the function \(\displaystyle f(z)=\frac{1}{(z+a)^4}\) which analytic and has poles at the point \(z=a\) of order 4.
Considering the square contour positively oriented having vertices \( \pm (1+i)\) and by residue lemma we derive ,
\( \displaystyle Res(a) = \frac{1}{6} \lim_{z\to a} \frac{d^3}{dz^3}(π\csc(πz))\)
Finally using alternate summation theorem ,
\(\displaystyle \sum_{n=\infty}^{\infty} \frac{(1)^n}{(n+a)^4} = \frac{1}{6} \lim_{z\to a} \frac{d^3}{dz^3}(π\csc(πz)) = \frac{\pi^4}{6} \frac{\cos(\pi a)}{\sin^2 (\pi a)} (6\csc^2 (\pi a) 1) \)
which gives the result as desired. – Aditya Narayan Sharma · 1 year ago
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– Aman Rajput · 1 year ago
firstly , i thought it can be solved using laurent series... but i thought we dont go to the complex analysis . :) niceLog in to reply
– Aditya Narayan Sharma · 1 year ago
Its nice to have both real and complex solutions of a problem for enlightenment as we now have for this oneLog in to reply
=\[\frac{1}{16} (\frac16(\psi^{(3)}(1\frac{a}2)+\psi^{(3)}(\frac{a}2)\psi^{(3)}(\frac{1a}2)\psi^{(3)}(\frac{1+a}2)))\]
Using reflection formula and On a little bit solving you get
\[\frac{\pi^4}{6}\frac{cos(\pi a)}{\sin^2(\pi a)}(6\csc^2(\pi a)1)\] Sorry for short solution – Aman Rajput · 1 year ago
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– Aditya Narayan Sharma · 1 year ago
Yup its ok. Nice solution .Log in to reply
Problem 21:
Prove that: \[\large \displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{n} = \sqrt{5}\]
This problem has been solved by Aditya Kumar. – Aditya Narayan Sharma · 1 year ago
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We use the well known generating function of Central Binomial coefficient:
\[\frac{1}{\sqrt{1  4x}} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}x^n\]
Here substitute \(x=\frac{1}{5}\).
Hence, \(\large \boxed{\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} 5^{n} = \sqrt{5}}\) – Aditya Kumar · 1 year ago
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– Harsh Shrivastava · 1 year ago
Proof of that generating function?Thanks.Log in to reply
here. – Aditya Kumar · 1 year ago
There are many ways but I learnt itLog in to reply
Problem 19 :
Prove That
\[ \sum_{r=1}^{n} H_{r}^{(2)} = (n+1)H_{n}^{(2)}  H_{n} \]
Notation : \(\displaystyle H_{n}^{(m)} = \sum_{k=1}^{n} \dfrac{1}{k^m} \) denotes the Generalized Harmonic Number.
This problem has been solved by Deeparaj Bhat. – Ishan Singh · 1 year ago
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– Deeparaj Bhat · 1 year ago
\[ \begin{align} \sum_{r=1}^n H_r^{(2)} &= \sum_{r=0}^{n1} \frac{nr}{(1+r)^2} \\&= \sum_{r=0}^{n1} \frac{n+1}{(1+r)^2}  \frac{1}{1+r} \\&= (n+1)H_n^{(2)}  H_n \\& Q. E. D. \end{align} \]Log in to reply
Problem 30 :
Evaluate:
\[\displaystyle \sum_{n=1}^{\infty} \frac{3^n}{\binom{2n}{n}} \] – Jack Lam · 1 year ago
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\[ f(x) = 2 \arcsin \left( \dfrac{x}{2} \right) = \sum_{n=1}^{\infty} \dfrac{x^n}{n^2 \dbinom{2n}{n}} \]
The series is \( \displaystyle x \dfrac{\text{d}}{\text{d}x} \left(x \dfrac{\text{d}}{{\text{d}x}} f(x) \right) \) at \(x=3\)
\( = 3+ \dfrac{4 \pi}{\sqrt{3}} \) – Ishan Singh · 1 year ago
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\[\displaystyle 9\int\limits_{0}^{1} \frac{x^42x^3+2x^2x}{(3x^23x+1)^2} dx\]
\[\frac{9}{27}(94\sqrt3\pi)\]
\[\boxed{\frac13(9+4\sqrt3\pi)}\] – Aman Rajput · 1 year ago
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Problem 28: Prove:
– Aman Rajput · 1 year agoLog in to reply
Let Jack post the next question. – Mark Hennings · 1 year ago
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– Aman Rajput · 1 year ago
This is called solution . Very nice sir ! Jusified clearlyLog in to reply
In the hopes it will help anyone, here is the decomposition.
\( \frac{1}{n^2 \binom{n+4}{4}} \equiv \frac{1}{n^2}  \frac{25}{12n} + \frac{4}{n+1}  \frac{3}{n+2} + \frac{4}{3(n+3)}  \frac{1}{4(n+4)} \) – Jack Lam · 1 year ago
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– Aditya Kumar · 1 year ago
You can proceed here by using the generating function.Log in to reply
\( \frac{(1)^{n+1}H_n}{n^2 \binom{n+4}{4}} \equiv \frac{(1)^n H_n}{n^2} + \frac{25}{12} \cdot \frac{(1)^n H_n}{n} + \frac{4(1)^{n+1} H_n}{n+1} + \frac{3(1)^{n+2}H_n}{n+2} + \frac{4}{3} \cdot \frac{(1)^{n+3} H_n}{n+3} + \frac{(1)^{n+4}}{4(n+4)} \)
Note the powers have been adjusted to match the denominator.
By using the Generating Function for the Harmonic Numbers and swapping the order of integration and summation, we obtain lots of logarithms, and a few Dilogarithms and Trilogarithms.
Use the Polylogarithm identities found on here: http://mathworld.wolfram.com/Trilogarithm.html and here: http://mathworld.wolfram.com/Dilogarithm.html to simplify everything. – Jack Lam · 1 year ago
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– Aman Rajput · 1 year ago
its better if you paste screenshots of wolframalpha.. that would help hereLog in to reply
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– Aman Rajput · 1 year ago
Hahaha ... 😂😂😂😂😂😂😂Log in to reply
– Jack Lam · 1 year ago
You asked for it, I presented it :)Log in to reply
– Aman Rajput · 1 year ago
i was not asking these .. i was asking the solution to the problem.Log in to reply
I do not know of a faster way to achieve the result. – Jack Lam · 1 year ago
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– Aman Rajput · 1 year ago
i know that the result you have posted and the screenshot does not match at all. I know this cant be solved using that. If you know then prove that you are right and i am wrong . If you cant then delete your question within a day. you have one more day to justify your resultLog in to reply
– Aman Rajput · 1 year ago
i also can add a solution to your problem using wolframalpha and put a screenshot there. But that wont be good.Log in to reply
– Aman Rajput · 1 year ago
you should post it . how we can believe your solution is correct.Log in to reply
– Aditya Kumar · 1 year ago
yes. That was one method. Posting all the steps is indeed tedious.Log in to reply
– Jack Lam · 1 year ago
I look forward to an elegant solution :)Log in to reply
Bonus Problem: Evaluate :
None has solved this question till 5 days – Aman Rajput · 1 year ago
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– Hung Woei Neoh · 12 months ago
...knock knock, is this comp over?Log in to reply
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– Hung Woei Neoh · 12 months ago
Apart from GPs and simple algebraic telescoping, I can't do summationsLog in to reply
\[\phi(e^{2n\pi})=\frac{kr_{2,2n^2}}{n^{1/4}2^{1/4}r_{n,n}}\]
Put \(n=6\) We have \[r_{6,6}=\frac{3^{1/8}\sqrt{1+\sqrt3}(1+\sqrt3+\sqrt23^{3/4})^{1/3}}{2^{13/24}}\]
\[r_{2,72}=\frac{\eta(6i)}{2^{1/4}\eta(12i)}\]
solving this function, and putting back into equation we get
– Aman Rajput · 12 months agoLog in to reply