How to calculate $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(m+n)!} $ ?

I don't know how to approach it . Please help :)

P.S.I am new to Double Sums and am not able to find any good sources to study it , can anyone help please ?

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    $\begingroup$ For each $j=1,2,3,\ldots$, how many $\frac{1}{j!}$ terms are there? $\endgroup$ – vadim123 Mar 9 '15 at 14:27
  • 1
    $\begingroup$ Apostol's Mathematical Analysis is a good source. $\endgroup$ – Tim Raczkowski Mar 9 '15 at 14:35
  • $\begingroup$ Perhaps start writing the sum out for a few terms. (Note that, since this is a double sum, you'll get a 2x2 array of numbers to add.) $\endgroup$ – Akiva Weinberger Mar 10 '15 at 15:56

This is the same as $$ \sum_{m=1}^\infty \sum_{k=m+1}^\infty \frac{1}{k!} $$ We can rearrange terms, noting that for each value of $k$ there will be terms only with $k > m$. There are $k-1$ possible values of $m$ that satisfy $k>m$. So $$ \sum_{m=1}^\infty \sum_{k=m+1}^\infty \frac{1}{k!} = \sum_{k=1}^\infty \frac{k-1}{k!} $$ The last trick is to note that it will be much easier to sum $\frac{k}{k!}$ so break up the numerator: $$ \sum_{k=1}^\infty \frac{k-1}{k!} = \sum_{k=1}^\infty \frac{k}{k!} - \sum_{k=1}^\infty \frac{1}{k!} = \sum_{k=1}^\infty \frac{1}{(k-1)!}- \sum_{k=1}^\infty \frac{1}{k!} $$ And this in turn is $$ \frac{1}{0!} + \sum_{k=2}^\infty \frac{1}{(k-1)!} - \sum_{k=1}^\infty \frac{1}{k!} = 1 +\sum_{j=1}^\infty \frac{1}{j!} - \sum_{k=1}^\infty \frac{1}{k!} $$ So far, only rearrangement of terms has happened. Now we note that $\sum_{k=1}^\infty \frac{1}{k!}$ is absolutely convergent, so the rearrangement of terms is valid; and the tow sums left cancel, so the answer is $$1 $$

  • $\begingroup$ Wow! You have brought it down to a single summation and solved it . Btw I have some doubts in your solution . k=m+1 , how did you make that substitution ? $\endgroup$ – Andy Mar 9 '15 at 15:50
  • $\begingroup$ Also where did n vanish to ? I know these maybe some foolish doubts , but sorry I am a novice to Double sums . Thanks $\endgroup$ – Andy Mar 9 '15 at 15:51
  • $\begingroup$ The substitution in the first line is $k=m+n$. That is where the $n$ vanished to. And since $k=m+n$ and $n>0$, that forces $k>m$. So the $k$ sum has to start at $k=m+1$. $\endgroup$ – Mark Fischler Mar 9 '15 at 16:26
  • $\begingroup$ One last doubt sir , how did the double sum get converted to a single summation . $\endgroup$ – Andy Mar 9 '15 at 16:31
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    $\begingroup$ @Andy: It became the numerator in $\frac{k-1}{k!}$, because $\frac{1}{k!}$ is a term in the summation for $k-1$ of the possible $m$s. $\endgroup$ – Henning Makholm Mar 9 '15 at 17:23

$$\sum_{m,n=1}^{+\infty}\frac{1}{(m+n)!}=\sum_{h\geq 2}\frac{f(h)}{h!}$$ where: $$ f(h)=\#\left\{(m,n)\in\mathbb{N}_{\geq 1}^2:m+n=h\right\}=h-1 $$ hence: $$\sum_{m,n=1}^{+\infty}\frac{1}{(m+n)!}=\sum_{h\geq 2}\frac{h}{h!}-\sum_{h\geq 2}\frac{1}{h!}=\sum_{h\geq 1}\frac{1}{h!}-\sum_{h\geq 2}\frac{1}{h!}=\color{red}{1}.$$



Under 'nice' conditions we have:


  • 1
    $\begingroup$ What's the nice condition , can you specify pls ? $\endgroup$ – Andy Mar 9 '15 at 16:31
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    $\begingroup$ @Andy: A sufficient condition is that the $a_k$s are all nonnegative. Then either the two sides of the equation are equal, or both sides diverge to $\infty$. $\endgroup$ – Henning Makholm Mar 9 '15 at 17:29


\begin{align} &\sum_{m=1}^\infty \sum_{n=1}^\infty \frac1{(m+n)!}\\ =& \sum_{n=1}^\infty \frac1{(1+n)!} + \sum_{n=1}^\infty \frac1{(2+n)!} + \sum_{n=1}^\infty \frac1{(3+n)!} + \dots\\ =&\vphantom{+}\frac1{2!} + \frac1{3!} + \frac1{4!} + \frac1{5!} + \dots\\ &+\frac1{3!} + \frac1{4!} + \frac1{5!} + \frac1{6!} + \dots\\ &+\frac1{4!} + \frac1{5!} + \frac1{6!} + \frac1{7!} + \dots \end{align}

Let's add and subtract $\sum\limits_{j=1}^\infty \frac1{j!}$. It doesn't change the sum and it's easier to add.

\begin{align*} =&\left(\frac1{1!}+\color{red}{\frac1{2!}}+\color{green}{\frac1{3!}}+\color{blue}{\frac1{4!}}+\dots\right.\\ &+\color{red}{\frac1{2!}}+\color{green}{\frac1{3!}}+\color{blue}{\frac1{4!}}+\frac1{5!}+\dots\\ &+\color{green}{\frac1{3!}}+\color{blue}{\frac1{4!}}+\frac1{5!}+\frac1{6!}+\dots\\ &+\color{blue}{\frac1{4!}}+\frac1{5!}+\frac1{6!}+\frac1{7!}+\dots\\ &\left.\vphantom{\frac12}+\dots\right)\\ &-\frac1{1!}-\frac1{2!}-\frac1{3!}-\frac1{4!}-\dots\\ &=\left(1+\color{red}{\frac1{1!}}+\color{green}{\frac1{2!}}+\color{blue}{\frac1{3!}}+\dots\right)-\left(\frac1{1!}+\frac1{2!}+\frac1{3!}+\dots\right)=\boxed{1} \end{align*}

Hope this helps! I know that the adding and subtracting $\sum\frac1{j!}$ came out of nowhere, but it works.

  • $\begingroup$ If anybody feels like replacing my picture with LaTeX, feel free to do so. $\endgroup$ – Akiva Weinberger Mar 10 '15 at 16:08
  • $\begingroup$ columbus8myhw: I have tried. Please, do check whether my edits correspond to what you wanted to write. $\endgroup$ – Martin Sleziak Mar 10 '15 at 16:48
  • $\begingroup$ @MartinSleziak Yeah, that's it. Thanks! $\endgroup$ – Akiva Weinberger Mar 10 '15 at 16:50

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