Hi Brilliant! I've decided to start the first ever summation contest here.
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:
1 2 3 4 5 6 7 |
|
The comments will be easiest to follow if you sort by "Newest":
See Part-2.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestProblem 2: Prove that n≥1∑n2(−1)n(γ+ψ(3n+21))=263ζ(3)+ζ(2)log2−14πG
G is Catalan's constant.
This problem has been solved by Ishan Singh.
Proposition:
Log in to reply
Solution Of Problem 2 :
Proof : It is easy to see that
n=1∑∞nHnxn=Li2(x)+21ln2(1−x)
Dividing by x and integrating, we have,
n=1∑∞n2Hnxn=∫xLi2(x)dx+21∫xln2(1−x)dx
=Li3(x)+21[ln2(1−x)lnx]+∫1−xlnxln(1−x)dx
Let x=1−t
⟹n=1∑∞n2Hnxn=Li3(x)+21[ln2(1−x)lnx]−∫tln(1−t)lntdt
Now,
∫tln(1−t)lntdt=lntLi2(t)−∫tLi2(t)dt
=lntLi2(t)−Li3(t)
Substituting back, we have,
n=1∑∞n2Hnxn=Li3(x)−Li3(1−x)+ln(1−x)Li2(1−x)+21lnxln2(1−x)+C
Putting x=0, we get C=ζ(3), thus,
n=1∑∞n2Hnxn=Li3(x)−Li3(1−x)+ln(1−x)Li2(1−x)+21lnxln2(1−x)+ζ(3)□
Now,
S=n=1∑∞n2(−1)n(γ+ψ(3n+21))=−2ln2n=1∑∞n2(−1)n+n=1∑∞k=1∑3n2k−12
Also,
k=1∑3n2k−11=k=1∑6nk1−k=1∑3n2k1
k=1∑3n2k−11=H6n−21H3n
⟹S=ζ(2)ln2+n=1∑∞n2(−1)n[2H6n−H3n]
Using cube roots and sixth roots of unity on the proposition, we have,
n=1∑∞n2(−1)n[2H6n−H3n]=263ζ(3)−14πG
⟹S=ζ(2)ln2+263ζ(3)−14πG
Log in to reply
SOLUTION OF PROBLEM 1 :
n≥0∑(−1)nψ(3)(2n+2)======3!n≥0∑(−1)nk≥0∑(k+2n+2)416n,k≥0∑(k+2n+2)4(−1)n=6K≥0∑⎝⎛n=0∑⌊K/2⌋(−1)n⎠⎞(K+2)416K≥0,K≡0,1(4)∑(K+2)4183K≥0∑(2K+1)41+6K≥0∑(4K+3)4183(1−161)901π4+1283ζ(4,23)2561π4+2561ψ(3)(43)
Log in to reply
Special Problem :
Evaluate
r=0∑n(rn)(−1)r
Log in to reply
Solution to Special Problem : Method 1
S=r=0∑n(rn)(−1)r
=r=0∑n(−1)rn!r!(n−r)!
=r=0∑n(−1)rn!r!(n−r)!(n−r+1+r+1)×(n+2)1
=n!(n+2)1r=0∑n[(−1)rr!(n−r+1)!+(−1)r(r+1)!(n−r)!]
=n!(n+2)1r=0∑n[(−1)rr!(n−r+1)!−(−1)r+1(r+1)!(n−r)!]
=n!(n+2)1r=0∑n(Tr−Tr+1)(*)
where Tr=(−1)rr!(n−r+1)!
Clearly, (∗) is a telescoping series. Evaluating it, we have,
S=(−1)nTn+1+T0
=(−1)nn!(n+2)(n+1)!+n!(n+2)(n+1)!
Since n is even,
∴S=2n+2n+1
Method 2
Consider the following Lemma.
Proof : S=r=0∑k(rk)(−1)r
=r=0∑k(k−rk)(−1)k−r
(∵r=0∑nf(r)=r=0∑nf(n−r))
=−r=0∑k(rk)(−1)r
⟹S=−S
⟹S=0
This completes the proof of the Lemma.
Now, since n is even, n+1 is odd.
Using the Lemma, we have,
r=0∑n+1(rn+1)(−1)r=0(1)
Let,
J=r=0∑n(rn)(−1)r(2)
Operating (1)+(2),
⟹J+0=r=0∑n(rn)(−1)r+r=0∑n+1(rn+1)(−1)r
=r=0∑n(rn)(−1)r+r=0∑n(rn+1)(−1)r+(n+1n+1)(−1)
=r=0∑n(rn)(−1)r+r=0∑n(n−rn+1)(−1)n−r−1(∵r=0∑nf(r)=r=0∑nf(n−r))
=r=0∑n(rn)(−1)r+r=0∑n(r+1n+1)(−1)r−1
=r=0∑n(−1)r⎩⎪⎪⎨⎪⎪⎧(rn)1+n+1r+1⋅(rn)1⎭⎪⎪⎬⎪⎪⎫−1[∵(rn)=rn(r−1n−1) & (rn)=(n−rn)]
⟹J=(n+1n+2)r=0∑n(rn)(−1)r+n+11r=0∑n(rn)(−1)r⋅r−1
⟹J+1=(n+1n+2)J+n+11= G (let)r=0∑n(rn)(−1)r⋅r
⟹J+1=(n+1n+2)J+n+1G(3)
Now,
G=r=0∑n(rn)(−1)r⋅r
=r=0∑n(n−rn)(−1)n−r⋅(n−r)
=r=0∑n(rn)(−1)r⋅(n−r)
=nr=0∑n(rn)(−1)r−r=0∑n(rn)(−1)r⋅r
⟹G=n⋅J−G
⟹G=2Jn(4)
From (3) & (4),
⟹J+1=(n+1n+2)J+2(n+1)Jn
∴J=2n+2n+1□
Log in to reply
Start by writing (rn)−1=(n+1)∫01ur(1−u)n−rdu.
Multiply both sides by (−1)r and adding over r gives:
r=0∑n(rn)(−1)r===(n+1)∫01r=0∑n(−u)r(1−u)n−rdu(n+1)∫01[(1−u)n+1+(−1)nun+1]dun+2n+1(1+(−1)n)=⎩⎨⎧0,n odd2n+4n+1,n even
Log in to reply
This is the same method which Ishan talked about. Thanks for posting!
Log in to reply
PROBLEM 11
Evaluate n=1∑∞(−1)n2n+1Hn.
This problem has been solved by Jake Lai and Aditya Kumar.
Log in to reply
It is well-known that −1−xln(1−x)=n=1∑∞Hnxn. So
n=1∑∞2n+1(−1)nHn=∫01n=1∑∞Hn(−x2)n dx=∫011+x2−ln(1+x2) dx=∫0π/4−ln(1+tan2u) du=∫0π/4−ln(sec2u) du=∫0π/42lncosu du=G−2πln2(Substitute u=tan−1x)
where G is Catalan's constant.
Log in to reply
Reversing the order of integration and summation to obtain the second line of your argument takes a bit of justification. The generating function series is not monotonic, nor is it uniformly bounded by an integrable function, so the result is not automatic.
You need to integrate from 0 to x<1, obtaining N=1∑∞2n+1(−1)nHnx2n+1=2∫0tan−1xlncosudu and then let x tend to 1−, using Abel's Theorem to get the answer (having used the Alternating Series Test to prove that the series ∑n(−1)n2n+1Hn is convergent).
Anyway, you are up for the next question!
Log in to reply
Alternate Method:
n=1∑∞2n+1(−1)nHn=n=1∑∞4n+12H2n−n=1∑∞2n+1Hn
Then we can use appropriate generating functions for evaluating it. The integrals have to be carefully handled in order to get the solution.
Log in to reply
Problem 3:
Evaluate :
r=0∑n[(rn)⋅sinrx⋅cos(n−r)x]
This problem has been solved by Deeparaj Bhat and Aditya Kumar.
Log in to reply
Call the given sum S.
Then, by using the definition of trigonometric functions via complex exponentials, we have S=4i1r=0∑n(rn)(einx−e−inx+e(2r−n)ix−e−(2r−n)ix)
Now, using binomial theorem and a bit of simplification, we get that S=2n−1sin(nx)
PS: I don't have good summation problems. So, anyone can post in my behalf. :)
Log in to reply
Alternate Solution:
We use the identity: r=0∑nf(r)=r=0∑nf(n−r)
We take:
S=r=0∑n[(rn)⋅sinrx⋅cos(n−r)x](1)
On applying the identity, we get
S=r=0∑n[(n−rn)⋅sin(n−rx)⋅cosrx](2)
On adding equations 1 and 2, we get:
2S=r=0∑n(rn)sinnx
Therefore, S=2n−1sinnx
Log in to reply
Problem 6 :
Prove That
r=1∑nr(−1)r−1(rn)=Hn
Notation : Hn denotes the Harmonic Number.
This problem has been solved by Aditya Kumar.
Log in to reply
Consider the binomial expansion:
r=0∑n(−x)r(rn)=(1−x)n
We'll write it as:
r=1∑n(−x)r(rn)=(1−x)n−1
On dividing both the sides by −x, we get:
r=1∑n(−x)r−1(rn)=x1−(1−x)n
On integrating both the sides w.r.t. x from 0 to 1, we get:
r=1∑nr(−1)r−1(rn)=∫01x1−(1−x)ndx=∫011−x1−xndx=Hn
Log in to reply
Solution To Problem 6 :
S=r=1∑nr(−1)r−1(rn)
Note that, k=r∑n(r−1k−1)=(rn), it follows,
S=r=1∑nr(−1)r−1k=r∑n(r−1k−1)
Since (r−1k−1)=kr(rk), we have,
S=r=1∑nk=r∑nk(−1)r−1(rk)
Interchanging order of summation, we have,
S=k=1∑nr=1∑kk(−1)r−1(rk)
=k=1∑nk1
=Hn□
Log in to reply
Problem 12:
Prove the following for ∣x∣<274.
n=0∑∞(n3n)xn=4−27x2cos(31sin−1(233x))
This problem has been solved by Mark Hennings and Ishan Singh.
Log in to reply
Solution to Problem 12:
By the multiplication formula for the Gamma function, (3n)!=Γ(3n+1)=2π132n+21Γ(n+31)Γ(n+32)Γ(n+1) and hence (n3n)=2π33n+21×Γ(2n+1)Γ(n+31)Γ(n+32)=2π32n+21B(n+32,n+31) Thus S=n=0∑∞(n3n)xn==2π3n=0∑∞32nxn∫01un−31(1−u)n−32du2π3∫011−27xu(1−u)u−31(1−u)−32du=2π3∫011−4sin23αu(1−u)u−31(1−u)−32du where 0<α<61π is such that sin3α=2127x. The substitutions u=sin2θ and then x=tanθ yield S===2π3∫021π1−sin23αsin22θsin−32θcos−34θ×2sinθcosθdθ=π3∫021π1−sin23αsin22θtan31θdθπ3∫0∞(1+x2)2−4sin23αx2x31(1+x2)dx=π3∫0∞(x2+2xsin3α+1)(x2−2xsin3α+1)x31(1+x2)dxπ3∫0∞x31f(x)dx=π3I where the meromorphic function f(z)=(z2+2zsin3α+1)(z2−2zsin3α+1)z2+1 has simple poles at each of the points in R={±ie3iα,±ie−3iα}.
Cutting the complex plane along the positive real axis, and integrating around the "keyhole contour" γ1+γ2−γ3−γ4 where
for any 0<ϵ<1<R, then (∫γ1+∫γ2−∫γ3−∫γ4)z31f(zdz=2πiu∈R∑Resz=uz31f(z) so that (1−ω)∫ϵRx31f(x)dx+(∫γ2−∫γ4)z31f(z)dz=2πiu∈R∑Resz=uz31f(z) Letting ϵ→0 and R→∞, we obtain (1−ω)I=2πiu∈R∑Resz=uz31f(z) and hence I=3(2cos2α−1)π=3(4cos2α−3)π=3cos3απcosα so that S=cos3αcosα=4−27x2cos(31sin−1(233x)) as required.
Log in to reply
Solution To Problem 12 :
Proof : Note that,
n=0∑∞Un(a)(−x)n=x2+2ax+11
where Un(x) is the Chebyshev Polynomial of the second kind.
⟹n=0∑∞Un(a)Γ(n+1)n!(−x)n=x2+2ax+11
Using Ramanujan Master Theorem, we have,
f(a,z)=sinπzπU−s(a)
=sinπzπsin(cos−1(a))sin((1−z)cos−1(a))□
Now, using Gamma Triplication Formula,
S=n=0∑∞(n3n)xn=2π3∫0∞(x2+2ax+1)(x2−2ax+1)x31(1+x2)dx
where a=233x
Using Partial Fraction and the Proposition, we have,
S=2π3[21f(31,−a)+21f(37,−a)−4a1f(34,−a)−4a1f(310,−a)+21f(31,a)+21f(37,a)+4a1f(34,a)+4a1f(310,a)]
After simplification, we have,
S=4−27x2cos(31sin−1(233x))□
Log in to reply
PROBLEM 13:
Show that n=1∑∞2n(Hn(2))2=3601π4−61π2ln22+61ln42+2Li4(21)+ζ(3)ln2
This problem has been solved by Aditya Kumar
Log in to reply
Solution to Problem 13:
Just to see how nasty the last integral in this problem can get, here are the associated indefinite integrals!
Start with B(x)===n=1∑∞(n+1)2Hn(2)xn+1=∫0x(∫0un=1∑∞Hn<