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

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

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TopNewestLet'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

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– Michael Tong · 3 years, 11 months ago

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.Log in to reply

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

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– Jimmy Kariznov · 3 years, 11 months ago

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.Log in to reply

– Pranav Arora · 3 years, 11 months ago

That makes sense, thank you.Log in to reply

– Jord W · 3 years, 10 months ago

that is incredible genius imo, you should be proud xDLog in to reply

– Alex Benfield · 3 years, 11 months ago

Forgive my ignorance but what is S(x)? Is it the same as f(x)?Log in to reply

– Jimmy Kariznov · 3 years, 11 months ago

\(S(x)\) is a function of \(x\) that I defined. I could have also named it \(f(x)\), or \(\xi(x)\), or even \(\clubsuit(x)\).Log in to reply

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

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

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\( \sum_{n = 1}^{\infty} \frac {1}{1 - x^n} - \frac{1}{1 - x^{n + 1}}\). 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

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Good explanation too !

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

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

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