\(\displaystyle \sum_{r=1}^{\infty} \frac{6^r}{(3^r - 2^r)(3^{r + 1} - 2^{r + 1})} \)
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\(\displaystyle \sum_{r=1}^{\infty} \frac{6^r}{(3^r - 2^r)(3^{r + 1} - 2^{r + 1})} \)
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Kushagraa Aggarwal
3 years, 11 months ago
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Top NewestLet'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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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
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Wow! Just great, Jimmy!
How did you even think of S(x)? – Pranav Arora · 3 years, 11 months ago
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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
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That makes sense, thank you. – Pranav Arora · 3 years, 11 months ago
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that is incredible genius imo, you should be proud xD – Jord W · 3 years, 10 months ago
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Forgive my ignorance but what is S(x)? Is it the same as f(x)? – Alex Benfield · 3 years, 11 months ago
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\(S(x)\) 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
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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
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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
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Thank you very much now I get it !
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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