\(\displaystyle\sum_{n=0}^{\infty} \frac{e}{3((n + 1)(n + 3))}\) \(=\) \(\frac{e}{4}\)
n=0∑∞3((n+1)(n+3))π = 4π
n=0∑∞3((n+1)(n+3))9 = 49
Therefore:
n=0∑∞3((n+1)(n+3))x will always equal 4x, where x is any type of number.
n=0∑∞(n+1)(n+3)x will always equal 43x, where x is any type of number.
Hope you found this interesting!
Calculus
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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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n=0∑∞(n+1)(n+3)1 = 21n=0∑∞(n+11−n+31)=43
So, n=0∑∞(n+1)(n+3)x will always be equal to 43x.
Also, a typo in the last summation, it should be 43x, not 12x
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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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