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

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

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

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!

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## Comments

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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

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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

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

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 $\frac{x}{12}$

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Ok, I will edit it. @Aaghaz Mahajan

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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.

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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!!!

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

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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.

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Look at my post to Aryan Sanghi.

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