# 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, 11 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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, 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.

- 6 years, 11 months ago

Wow! Just great, Jimmy!

How did you even think of S(x)?

- 6 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.

- 6 years, 11 months ago

That makes sense, thank you.

- 6 years, 11 months ago

How will we get

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

- 6 years, 11 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, $\lim_{n \rightarrow \infty} - \frac{1}{1 - x^{n + 1}} = - \frac {1}{1 - 0}$.

- 6 years, 11 months ago

Thank you very much now I get it !

Good explanation too !

Thank You !

- 6 years, 11 months ago

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

- 6 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 $\clubsuit(x)$.

- 6 years, 11 months ago

that is incredible genius imo, you should be proud xD

- 6 years, 10 months ago

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

- 6 years, 11 months ago