Everybody knows the following infinite series :

\[\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!} = e-1 \]

which comes out from the Taylor Series of \(e^x\). It is a single summation.

What about the following :

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

Or, Let's generalize that for \(n\) summations :

\[\displaystyle \sum_{a_0,a_1,a_2,\dots,a_{n-1} >0 } \frac{1}{(a_0+a_1+a_2+\dots+a_{n-1})!} \]

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

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

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

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

\[\displaystyle \sum_{a_0,a_1,a_2,\dots,a_{n-1} >0 } \frac{1}{(a_0+a_1+a_2+\dots+a_{n-1})!} \]

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

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

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TopNewestNice question.

Here's my approach:

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

Focus on the denominator \(a_{0}+a_{1}+a_{2}...a_{n-1}\).

Fix it's value to be some constant \(k\) (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 \[a_{0}+a_{1}+a_{2}...a_{n-1}=k\]

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

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

Note also, that the smallest possible value of \(k\) is n and it goes on up until infinity.

Thus, our summation can be written as:

\[S(n)=\sum_{k=n}^{\infty}\frac{\binom{k-1}{n-1}}{k!}\]

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

For convenience, let \(k-n=r\).

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

We'll deal with the \((n-1)!\) later. \[\]

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

\[\] Also, \(\quad \quad \quad e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}\).

\[\Rightarrow x^{n-1}e^x=\sum_{r=0}^{\infty}\frac{x^{r+n-1}}{r!}\]

\[\]

\[\Rightarrow \int_{0}^{1}x^{n-1}e^xdx=\int_{0}^{1}\sum_{r=0}^{\infty}\frac{x^{r+n-1}}{r!}dx\]

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

Thus,

\[T(n)=\int_{0}^{1}x^{n-1}e^xdx\]

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

\[\]

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

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Very elegant solution!

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

Nice work :)

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Thanks :)

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

I've edited it now.

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