Interesting Thing about Infinite Series...

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

n=0π3((n+1)(n+3))\displaystyle\sum_{n=0}^{\infty} \frac{\pi}{3((n + 1)(n + 3))} == π4\frac{\pi}{4}

n=093((n+1)(n+3))\displaystyle\sum_{n=0}^{\infty} \frac{9}{3((n + 1)(n + 3))} == 94\frac{9}{4}


n=0x3((n+1)(n+3))\displaystyle\sum_{n=0}^{\infty} \frac{x}{3((n + 1)(n + 3))} will always equal x4\frac{x}{4}, where xx is any type of number.

n=0x(n+1)(n+3)\displaystyle\sum_{n=0}^{\infty} \frac{x}{(n + 1)(n + 3)} will always equal 3x4\frac{3x}{4}, where xx is any type of number.

Hope you found this interesting!



Note by A Former Brilliant Member
4 weeks ago

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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

A Former Brilliant Member - 1 week, 1 day ago

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n=01(n+1)(n+3) = 12n=0(1n+11n+3)=34\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, n=0x(n+1)(n+3)\sum_{n=0}^{\infty}\frac{x}{\left(n+1\right)\left(n+3\right)} will always be equal to 3x4\frac{3x}{4}.

Also, a typo in the last summation, it should be 3x4\frac{3x}{4}, not x12\frac{x}{12}

Aaghaz Mahajan - 4 weeks ago

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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.

Aryan Sanghi - 4 weeks ago

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Hey man, are you the younger brother of Aniket Sanghi?? i used to follow his feed for those amazing doubt he secured a great rank in JEE!!!

Aaghaz Mahajan - 4 weeks ago

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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.

Aruna Yumlembam - 4 weeks ago

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Look at my post to Aryan Sanghi.

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