\(\displaystyle\sum_{n=0}^{\infty} \frac{e}{3((n + 1)(n + 3))}\) \(=\) \(\frac{e}{4}\)

$\displaystyle\sum_{n=0}^{\infty} \frac{\pi}{3((n + 1)(n + 3))}$ $=$ $\frac{\pi}{4}$

$\displaystyle\sum_{n=0}^{\infty} \frac{9}{3((n + 1)(n + 3))}$ $=$ $\frac{9}{4}$

Therefore:

$\displaystyle\sum_{n=0}^{\infty} \frac{x}{3((n + 1)(n + 3))}$ will always equal $\frac{x}{4}$, where $x$ is any type of number.

$\displaystyle\sum_{n=0}^{\infty} \frac{x}{(n + 1)(n + 3)}$ will always equal $\frac{3x}{4}$, where $x$ is any type of number.

Hope you found this interesting!

No vote yet

1 vote

Easy Math Editor

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:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@Alak Bhattacharya, @Mahdi Raza, @Richard Desper, @Gandoff Tan, @Hamza Anushath, @Adhiraj Dutta, @Alice Smith, @Isaac YIU Maths Studio, @Chew-Seong Cheong, @Páll Márton, @Darsh Kedia, @Vinayak Srivastava, @Zakir Husain, @Pradeep Tripathi, @Aryan Sanghi, @Culver Kwan, @Pi Han Goh, @Shikhar Srivastava, @Aaghaz Mahajan, @Sahar Bano, @Karan Chatrath, @Sachetan Debray

Log in to reply

You were right! You do know more people. XD. But I don't know a thing about Calculus, Limits or Integrals, so I don't understand the note. =D

Log in to reply

$\sum_{n=0}^{\infty}\frac{1}{\left(n+1\right)\left(n+3\right)}\ =\ \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)=\frac{3}{4}$

So, $\sum_{n=0}^{\infty}\frac{x}{\left(n+1\right)\left(n+3\right)}$ will always be equal to $\frac{3x}{4}$.

Also, a typo in the last summation, it should be $\frac{3x}{4}$, not $\frac{x}{12}$

Log in to reply

Ok, I will edit it. @Aaghaz Mahajan

Log in to reply

Nice post @Yajat Shamji. Thanku for sharing this with us.

Alternatively, you could go from your result to observation and proof it like @Aaghaz Mahajan mentioned . Found your post interesting.

Log in to reply

Hey man, are you the younger brother of Aniket Sanghi?? i used to follow his feed for those amazing problems....no doubt he secured a great rank in JEE!!!

Log in to reply

Thanks, @Aryan Sanghi! But I am GCSE so am only posting this for comments and suggestions

Log in to reply

Can you please send a proof ? Also you can prove more sums like this via the Digamma function or atleast that's how I solved your sums.Try Brilliant website for the Digamma function.

Log in to reply

Look at my post to Aryan Sanghi.

Log in to reply

Knowing the convergence of the series, this result is trivial if the sum can be calculated.

Log in to reply