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

Note by Kushagraa Aggarwal
4 years, 5 months ago

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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:

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

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

- 4 years, 5 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.

- 4 years, 5 months ago

Wow! Just great, Jimmy!

How did you even think of S(x)?

- 4 years, 5 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.

- 4 years, 5 months ago

That makes sense, thank you.

- 4 years, 5 months ago

that is incredible genius imo, you should be proud xD

- 4 years, 4 months ago

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

- 4 years, 5 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 $$\clubsuit(x)$$.

- 4 years, 5 months ago

How will we get

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

- 4 years, 5 months 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, $$\lim_{n \rightarrow \infty} - \frac{1}{1 - x^{n + 1}} = - \frac {1}{1 - 0}$$.

- 4 years, 5 months ago

Thank you very much now I get it !

Good explanation too !

Thank You !

- 4 years, 5 months ago

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

- 4 years, 5 months ago