Hi, everyone!

Today, I found an amazing new identity, and I wish to share with you all. Well, it states

\[\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{x^n}{m{n+m \choose m}} = \log \left(\frac{1}{1-x}\right)\]

I have added a wiki on it - Kishlaya's Identity

Since, I was very excited that I didn't thought about having a new name for it. So, do you have some suggestions for name?

All kind of feedbacks will be appreciated.

Thanks,

Kishlaya Jaiswal

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## Comments

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TopNewestHere is a simple proof of the identity. First, we write \[\sum_{n = 1}^\infty \sum_{m = 1}^\infty \frac{x^n}{m \binom{n + m}{m}} = \sum_{n = 1}^\infty x^n \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!}.\]

Then \[ \begin{align*} \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!} &= \sum_{m = 1}^\infty \frac{(m - 1)! n!}{(m + n)!} \\ &= \frac{1}{n} \sum_{m = 1}^\infty \frac{(m - 1)! n! \cdot n}{(m + n)!} \\ &= \frac{1}{n} \sum_{m = 1}^\infty \frac{(m - 1)! n! \cdot [(m + n) - m]}{(m + n)!} \\ &= \frac{1}{n} \sum_{m = 1}^\infty \left( \frac{(m - 1)! n! \cdot (m + n)}{(m + n)!} - \frac{(m - 1)! n! \cdot m}{(m + n)!} \right) \\ &= \frac{1}{n} \sum_{m = 1}^\infty \left( \frac{(m - 1)! n!}{(m + n - 1)!} - \frac{m! n!}{(m + n)!} \right). \end{align*} \] This sum telescopes to \(1/n\), so \[\sum_{n = 1}^\infty x^n \sum_{m = 1}^\infty \frac{m! n!}{m (m + n)!} = \sum_{n = 1}^\infty \frac{x^n}{n} = \log \frac{1}{1 - x}.\]

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This is definitely amazing.......Genius!!!

I'll definitely add this proof (by your name) in my article and wiki.

Thanks.

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Kishlaya , the name's fine. Not everyone is lucky enough to have an identity with their name attached to it .

Congrats \(\ddot\smile\)

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Oh, thank you so much!!!

Actually, I wanted to have some appealing and striking name but I'd no idea for the same.

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Kishlaya's a beautiful name! Suiting for a beautiful identity.

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this .

Right !!!, seeLog in to reply

@Azhaghu Roopesh M You're truly dynamic, restless, independent, ready to accept challenges, and outspoken. You enjoy change, travel, and new experiences.

And this is far more better.

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@Jake Lai I am sure that soon I'll be seeing another beautiful identity named after a beautiful name Jake Lai, the great mathematician .

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What is the geometric meaning of double summation?

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Hey, can you please tell me where can I learn double and triple summation from ? Or can you please write a wiki on it here on Brilliant ?

I've searched a lot of places but got no satisfactory info .

Thanks for the same .

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Try studying a bit of discrete calculus.

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This is a pretty good source, and covers the difference, sum, and rules pretty thoroughly. Kishlaya's own identity uses the Beta function, though, rather than discrete methods.

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I am too a learner

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I wish I had as much brains as you do!

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Indeed, you have much more brain than me.

Getting a new identity doesn't makes me more intelligent than you. :)

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Let's start from basics(I am confused right now)

What's the definition of summation and integral according to you?

How can you simultaneously works with 2 variables?

What does the upper limit as infinity means?

i am unable to understand the logic behind interchanging integral and sum ?

How can you consider the summation part as a slope for the integral?

I am totally confused seeing 2 sigma right now(what do they really mean , why are you taking it independent?) Can I consider infinite sigma with infinite integral such as( a reverse picture of what you were having i.e summand upper limit infinity and integgral upper limit as n) \( \displaystyle \sum_{a}^{\infty} \sum_{b}^{\infty} \cdots \int_{0}^{n} da \int_{0}^{n} db \cdots = \int_{0}^{n} \sum_{a}^{\infty} da \int_{0}^{n} \sum_{b}^{\infty}db \cdots\) ?

At last what is calculus according to you(I think I understood calculus in a wrong manner)?

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Well, I got what you are confused with.

Actually, it's just a special case of Fubini/Tonelli theorems.

Whereas I'll try my best to update that wiki to completely clear your doubts but in the meantime you can look up here

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Consider this :

\(\displaystyle \int { f(x)dx } +\int { g(x)dx } = \int { f(x)+g(x)dx } \)

What does this mean. This means that you have evaluating an expression in two ways, you first integrate the functions seperately and than add it, or rather you add the two functions to get the new funciton and then integrate it. You could do it with 3 functions as follows :

\(\displaystyle \int { f(x)dx } +\int { g(x)dx } +\int { h(x)dx } =\int { f(x)+g(x)+h(x)dx } \)

Or with \(n\) functions :

\(\displaystyle \int { { f }_{ 1 }(x)dx } +\int { { f }_{ 2 }(x)dx } +.....+\int { { f }_{ n }(x)dx } =\int { { f }_{ 1 }(x)+{ f }_{ 2 }(x)+.....{ f }_{ n }(x)dx } \)

So if you write that in terms of sigma notation you have :

\(\displaystyle \sum _{ r=1 }^{ n }{ \int { { f }_{ r }(x)dx } } =\int { \sum _{ r=1 }^{ n }{ { f }_{ r }(x) } dx } \)

This is called interchange of summation and integral.

I hope this clear things a bit.

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Yes , yes thank you at that time I was confused to see double summation with integral , the after sometime I understood it. OK nice explanation thank you.

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there r already so many names

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This identity only seems to hold for \(x>-1\)?

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Yes Sir, I mentioned that in the wiki.

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Oh there it is, I see it. I have to admit, that's quite an awesome identity. And it's one of those fascinating cases where 2 different functions can be identical over a certain range, and suddenly go separate ways.

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Well, I'm researching over it yet, because I still think, there may exist some other interesting variant.

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generalized version of this identity may amaze you more.

Sir, then theLog in to reply

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Sir, that's just too great of you and thank you so much.

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MathWorld

Wolfram Research, Inc.

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Champaign, IL 61820-7237

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fax: +1-217-398-0747

and then there's Wolfram Special Functions site, which is more tightly limited, but will still consider contributions of interest, and may be contacted at:

comments@functions.wolfram.com

Do not contact "WolframAlpha", you really should start with www.wolfram.com and go from there. It' s a pretty big organization, so you'll need patience in finding the right contact interested in your work. Understand that even if Wolfram decides not to publish your identity, they still may want to use it as part of their "knowledge base". Wolfram Mathematica relies on voluntary contributions from a large base of mathematicians.

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Thanks.

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@Kishlaya Jaiswal

Best of LuckLog in to reply

Can you suggest what to do next because I am not sure if a demonstration can still be made out of that identity?

[P.S. - In the meantime, I've found two other interesting identities/series.]

Thanks.

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can u plse tell all of us the proof

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Yep, you can read the complete article/wiki (proof) here - Kishlaya's Identity

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