Interesting Thing about Infinite Series...

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

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

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

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!

Calculus

Algebra

#Calculus

Easy Math Editor

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Comments

@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

A Former Brilliant Member - 1 year ago

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

A Former Brilliant Member - 11 months, 3 weeks ago

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

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 $\frac{x}{12}$

Aaghaz Mahajan - 1 year ago

Ok, I will edit it. @Aaghaz Mahajan

A Former Brilliant Member - 1 year ago

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.

Aryan Sanghi - 1 year ago

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!!!

Aaghaz Mahajan - 1 year ago

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

A Former Brilliant Member - 1 year ago

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.

Aruna Yumlembam - 1 year ago

Look at my post to Aryan Sanghi.

A Former Brilliant Member - 1 year ago

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

William Ly - 4 months, 4 weeks ago

Math	Appears 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}`	$\boxed{123}$