# Interesting Infinite Sums

Here are some interesting values of infinite sums that I calculated:

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)} = \frac{\pi}{3\sqrt{3}}$

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+3)} = \frac{\pi}{8}$ (Leibniz's formula)

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(6n+1)(6n+5)} = \frac{\pi}{8\sqrt{3}}$

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(8n+1)(8n+7)} = \frac{\pi}{48}(\sqrt{2}+1)$

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(8n+3)(8n+5)} = \frac{\pi}{16}(\sqrt{2}-1)$

My method for calculation using the result:

$\displaystyle \sum_{r=1}^n \cos rx = \frac{\sin(n+\frac{1}{2})x-\sin\frac{1}{2}x}{2\sin\frac{1}{2}x}$ for $\sin\frac{1}{2}x\neq0$

Integrating both sides we get $\displaystyle \sum_{r=1}^n \frac{1}{r}\sin rx = \int \frac{\sin(n+\frac{1}{2})x-\sin\frac{1}{2}x}{2\sin\frac{1}{2}x} dx$

Simplifying the integral as making the appropriate substitution $u=\frac{1}{2}x$ we get the result:

$x + \displaystyle \sum_{r=1}^n \frac{1}{r}\sin 2rx = \int \frac{\sin (2n+1)x}{\sin x} dx$

Using the fact that $\lim_{n\to\infty} \int_{0}^{a} \frac{\sin (2n+1)x}{\sin x} dx = \frac{\pi}{2}$ for positive $a$ and $a\neq q\pi$ where $q$ is a positive integer, we can establish infinite sums such as:

$\frac{\pi}{6} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{3} = \frac{\pi}{2}$

$\frac{\pi}{8} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{4} = \frac{\pi}{2}$

$\frac{\pi}{3} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{2r\pi}{3} = \frac{\pi}{2}$

$\frac{\pi}{4} + \displaystyle \sum_{r=1}^{\infty} \frac{1}{r}\sin \frac{r\pi}{2} = \frac{\pi}{2}$

Note by Chris Sapiano
4 months, 3 weeks 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:

Niceee!!! I even liked solving your recent problems!!!

- 4 months, 3 weeks ago

Hey just wanted to know that the above 5 sums you calculated by the Digamma function or not.If not please send me a different solution .Also I proved all 5 sums via digamma function.

- 3 months, 3 weeks ago

No i wrote how i proved it underneath

- 3 months, 3 weeks ago

Thanks,well now I can find some new results regarding Digamma function.

- 3 months, 3 weeks ago

Can you show them? I have no idea what the digamma function is

- 3 months, 3 weeks ago

The Digamma function you see is simply the logarithmic derivative of the gamma function that is take the natural log of the gamma function and take its derivative respect to the input.You can check the more clear definition at Brilliant website .You can also look at my discussion at the calculus section titled 'On the Iterated Properties of Digamma function'.Also it's one of my favourite function.

- 3 months, 3 weeks ago