The Mighty Taylor Series of ee.

Everybody knows the following infinite series :

n=11n!=e1\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!} = e-1

which comes out from the Taylor Series of exe^x. It is a single summation.

What about the following :

n=1m=11(n+m)!\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)!}

Or, Let's generalize that for nn summations :

a0,a1,a2,,an1>01(a0+a1+a2++an1)!\displaystyle \sum_{a_0,a_1,a_2,\dots,a_{n-1} >0 } \frac{1}{(a_0+a_1+a_2+\dots+a_{n-1})!}

Does this have a closed form? It is a challenge for you guys!

Challenge  1  :(Easy)\mathbf{Challenge \; 1\; :} {\color{#20A900}{\textrm{(Easy)}}} : Find the value of :

n=1m=11(n+m)!\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)!}

Challenge  2  :(Hard)\mathbf{Challenge \; 2\; :} {\color{#D61F06}{\textrm{(Hard)}}} : Find the closed form of :

a0,a1,a2,,an1>01(a0+a1+a2++an1)!\displaystyle \sum_{a_0,a_1,a_2,\dots,a_{n-1} >0 } \frac{1}{(a_0+a_1+a_2+\dots+a_{n-1})!}

Try them ! I will share my approach the next week.

This note is a part of the set Sequences and Series Challenges.

Note by Hasan Kassim
5 years ago

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1 vote

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Nice question.

Here's my approach:

Denote the required sum by S(n)S(n).

Focus on the denominator a0+a1+a2...an1a_{0}+a_{1}+a_{2}...a_{n-1}.

Fix it's value to be some constant kk (for now). Consider all the terms in our summation with this denominator. How many such terms are there?

That's easy: the number of such terms is simply the number of solutions to the equation a0+a1+a2...an1=ka_{0}+a_{1}+a_{2}...a_{n-1}=k

Note that we need solutions in the natural numbers only as all the aia_{i} vary only from 1 to infinity.

So, for a given k, the number of such terms is given by (k1n1)\binom{k-1}{n-1}.

Note also, that the smallest possible value of kk is n and it goes on up until infinity.

Thus, our summation can be written as:


After simplification, this can be written as: S(n)=1(n1)!k=n1k(kn)!S(n)=\frac{1}{(n-1)!}\sum_{k=n}^{\infty}\frac{1}{k(k-n)!}

For convenience, let kn=rk-n=r.

Thus, S(n)=1(n1)!r=01(r+n)(r)!S(n)=\frac{1}{(n-1)!}\sum_{r=0}^{\infty}\frac{1}{(r+n)(r)!}

We'll deal with the (n1)!(n-1)! later.

For now, let T(n)=r=01(r+n)(r)!\quad \quad T(n)=\sum_{r=0}^{\infty}\frac{1}{(r+n)(r)!}

Also, ex=r=0xrr!\quad \quad \quad e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}.

xn1ex=r=0xr+n1r!\Rightarrow x^{n-1}e^x=\sum_{r=0}^{\infty}\frac{x^{r+n-1}}{r!}

01xn1exdx=01r=0xr+n1r!dx\Rightarrow \int_{0}^{1}x^{n-1}e^xdx=\int_{0}^{1}\sum_{r=0}^{\infty}\frac{x^{r+n-1}}{r!}dx

Note that the RHS is the expression that we had obtained earlier for T(n)T(n).



Integrate this by parts to establish the recurrence relation: T(n)=e(n1)T(n1)T(n)=e-(n-1)T(n-1)

This is a recurrence relation. Solve it and you're done. Once you've found T(n)T(n), divide that by (n1)!(n-1)! to get S(n)S(n).

Shashwat Shukla - 5 years ago

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Very elegant solution!

Just a thing: the expression you obtained from the integral for T(n)T(n) should be for T(n+1)T(n+1) instead.

Nice work :)

Hasan Kassim - 5 years ago

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Thanks :)

Really enjoyed doing this problem....And yea, it should be T(n+1)T(n+1).

I've edited it now.

Shashwat Shukla - 5 years ago

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@Shashwat Shukla Also replace the coefficient of T(n1)T(n-1) by n1n-1, in the recursion.

Hasan Kassim - 5 years ago

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@Hasan Kassim My bad. Done now...Oh and btw, I saw this post when I was going through the problems of your set, 'Sequences and series challenges' and I liked all of your problems. Thanks for posting them :) Cheers

Shashwat Shukla - 5 years ago

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@Shashwat Shukla Yea I will keep posting new ones, my pleasure :)

Hasan Kassim - 5 years ago

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Its quite tough.

Ayush Kumar - 1 year, 10 months ago

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