Waste less time on Facebook — follow Brilliant.
×

I once submitted this question .

\(\displaystyle \sum_{r=1}^{\infty} \frac{6^r}{(3^r - 2^r)(3^{r + 1} - 2^{r + 1})} \)

Note by Kushagraa Aggarwal
3 years, 11 months ago

No vote yet
11 votes

Comments

Sort by:

Top Newest

Let's try to generalize. For \(|x| < 1\) we have the following:

\(S(x) := \displaystyle \sum_{n = 1}^{\infty} \dfrac{x^n}{(1-x^n)(1-x^{n+1})}\) \(= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty} \dfrac{x^n(1-x)}{(1-x^n)(1-x^{n+1})}\)

\(= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty} \dfrac{x^n - x^{n+1}}{(1-x^n)(1-x^{n+1})}\) \(= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty} \dfrac{(1- x^{n+1}) - (1-x^n)}{(1-x^n)(1-x^{n+1})}\)

\(= \dfrac{1}{1-x}\displaystyle \sum_{n = 1}^{\infty}\left[\dfrac{1}{1-x^n} - \dfrac{1}{1-x^{n+1}}\right]\) \(= \dfrac{1}{1-x}\left[ \dfrac{1}{1-x^1} - \dfrac{1}{1-0}\right]\) \(= \dfrac{x}{(1-x)^2}\).

Now to tackle the sum in question:

\(\displaystyle\sum_{n = 1}^{\infty} \dfrac{6^n}{(3^n-2^n)(3^{n+1}-2^{n+1})}\) \(= \displaystyle\sum_{n = 1}^{\infty} \dfrac{3^{2n}\left(2/3\right)^n}{3^{2n+1}\left(1-\left(2/3\right)^n\right)\left(1-\left(2/3\right)^{n+1}\right)}\)

\(= \dfrac{1}{3}\displaystyle\sum_{n = 1}^{\infty} \dfrac{\left(2/3\right)^n}{\left(1-\left(2/3\right)^n\right)\left(1-\left(2/3\right)^{n+1}\right)}\) \(= \dfrac{1}{3}S\left(\dfrac{2}{3}\right)\) \(= \dfrac{1}{3} \dfrac{2/3}{\left(1-2/3\right)^2} = \boxed{2}\) Jimmy Kariznov · 3 years, 11 months ago

Log in to reply

@Jimmy Kariznov The best part of this solution is that it's deceptively straight forward. If somebody were to see all of these "fancy" math symbols and such they would instinctively feel intimidated, but if you really look at what you're doing here it's really just very smart equation manipulation. Michael Tong · 3 years, 11 months ago

Log in to reply

@Jimmy Kariznov Wow! Just great, Jimmy!

How did you even think of S(x)? Pranav Arora · 3 years, 11 months ago

Log in to reply

@Pranav Arora When solving this question, I first did the manipulations to the summation. I then replaced \(2/3\) with \(x\) to see if there was some clever power series manipulations that could be done. As it turned out, it was just a telescoping sum, and replacing \(2/3\) with \(x\) was not really necessary. However, \(x\) is easier to type repeatedly than \(2/3\), and so, I showed the generalization. Jimmy Kariznov · 3 years, 11 months ago

Log in to reply

@Jimmy Kariznov That makes sense, thank you. Pranav Arora · 3 years, 11 months ago

Log in to reply

@Jimmy Kariznov that is incredible genius imo, you should be proud xD Jord W · 3 years, 10 months ago

Log in to reply

@Jimmy Kariznov Forgive my ignorance but what is S(x)? Is it the same as f(x)? Alex Benfield · 3 years, 11 months ago

Log in to reply

@Alex Benfield \(S(x)\) is a function of \(x\) that I defined. I could have also named it \(f(x)\), or \(\xi(x)\), or even \(\clubsuit(x)\). Jimmy Kariznov · 3 years, 11 months ago

Log in to reply

@Jimmy Kariznov How will we get

\( \sum_{n=1}^{\infty} \Bigg[ -\frac{1}{1-x^{n+1}} \Bigg] = -\frac{1}{1-0} \) ?

Please help , Thank You Priyansh Sangule · 3 years, 11 months ago

Log in to reply

@Priyansh Sangule Look at the whole expression. It states

\( \sum_{n = 1}^{\infty} \frac {1}{1 - x^n} - \frac{1}{1 - x^{n + 1}}\). So, when \(n = 1\) then you have \( \frac {1}{1 - x} - \frac {1}{1 - x^2}\). Add \(n = 2\) you have \( \frac {1}{1 - x^2} - \frac {1}{1 - x^3}\). Notice how the first term of this cancels with the second term of the first. This is called a telescopic series. Thus, you will only have the first term ( \( \frac{1}{1 - x}\)) and the last term, \( \lim_{n \rightarrow \infty} - \frac{1}{1 - x^{n + 1}} = - \frac {1}{1 - 0} \). John Cain · 3 years, 11 months ago

Log in to reply

@John Cain Thank you very much now I get it !

Good explanation too !

Thank You ! Priyansh Sangule · 3 years, 11 months ago

Log in to reply

If I'm not mistaken, this was a Putnam problem! Jess Smith · 3 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...