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
6 years, 1 month ago

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If I'm not mistaken, this was a Putnam problem!

- 6 years, 1 month ago

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:

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

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

- 6 years, 1 month ago

that is incredible genius imo, you should be proud xD

- 6 years ago

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

- 6 years, 1 month ago

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

- 6 years, 1 month 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.

- 6 years, 1 month ago

How will we get

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

- 6 years, 1 month ago

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, $\lim_{n \rightarrow \infty} - \frac{1}{1 - x^{n + 1}} = - \frac {1}{1 - 0}$.

- 6 years, 1 month ago

Thank you very much now I get it !

Good explanation too !

Thank You !

- 6 years, 1 month ago

Wow! Just great, Jimmy!

How did you even think of S(x)?

- 6 years, 1 month 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.

- 6 years, 1 month ago

That makes sense, thank you.

- 6 years, 1 month ago