2
$\begingroup$

I am working with the series

$$\sum_{n,m = 1}^{\infty} \frac{1}{\left(n+m\right)!}$$

and I would like to show that it converges absolutely. It is easy to see that it actually converges to $1$ as

$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{\left(n+m\right)!} = \sum_{k=2}^{\infty}\sum_{n=1}^{k-1}\frac{1}{k!} = \sum_{k=2}^{\infty}\left(\frac{1}{(k-1)!} - \frac{1}{k!} \right) = 1 $$

but does that consitute a formal proof? I thought I could instead show that the partial sums are bounded as in that case the nonnegativity would imply convergence but I have not been able to do that. Could you please give me some advice? Thank you.

$\endgroup$
  • $\begingroup$ where is $n$? By the other side I have that $\sum \frac1{(n+m)!}=\sum \frac{k-1}{k!}$ that is equal to your proof. In other words: it must be correct :p $\endgroup$ – Masacroso Oct 28 '16 at 22:16
  • $\begingroup$ @Masacroso $n$ is a dummy, make the substitutions $k=(n+m)$ and $n=n$, then you will have that $n$ ranges from $1$ to $k-1$ since $m$ is larger than or equal to $1$. $\endgroup$ – JohnK Oct 28 '16 at 22:25
  • 1
    $\begingroup$ @Masacroso It's not a mistake, this is what happens after the substitution: there is no n. $\endgroup$ – JohnK Oct 28 '16 at 22:31
  • 2
    $\begingroup$ Yes, when all terms that you're summing are positive, all of these manipulations are valid. $\endgroup$ – Alex Zorn Oct 28 '16 at 22:38
  • $\begingroup$ @AlexZorn indeed if they were not positive numbers then the notation index $m,n=1$ is an abuse of notation. $\endgroup$ – Masacroso Oct 28 '16 at 22:50
1
$\begingroup$

The given answers assert that this double series converges to the value obtained by summing diagonally:

$$\sum_{n,m \in \mathbb{N}} \frac{1}{(n+m)!} = \sum_{n =2}^\infty \sum_{j+k = n} \frac{1}{(j+k)!} = \sum_{n=2}^\infty\frac{n-1}{n!}= 1.$$

This is indeed true since the terms are non-negative. In this case, the double series converges if and only if sums by rows or columns (iterated series), diagonal sums, or any other arrangement converge and all such sums will be equal. This fact is so well known that it is often taken for granted. However, this result is not generally true for conditionally convergent series where terms change sign.

For the formal proof you asked about, consider the general treatment of double series $\sum_{m,n} a_{mn}$ with $a_{mn} > 0$.

We define the partial sum

$$S_{mn} = \sum_{j=1}^m \sum_{k=1}^n a_{jk},$$

and a diagonal sum

$$\sigma_n = \sum_{p=2}^n \sum_{j+k= p}a_{jk}.$$

We say that the double series converges to $S$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $m,n \geqslant N$ implies $|S_{mn} -S| < \epsilon.$

First, the double series converges to $S$ if and only if we can sum by squares:

$$S = \lim_{n \to \infty}S_{nn}.$$

The forward implication is easily proved by observing that with $m=n \geqslant N$ we have $|S_{nn} -S| < \epsilon.$

For the reverse implication, we see that if $S_n \to S$, then given $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n \geqslant N$ implies $S- \epsilon < S_{nn} \leqslant S.$ Thus, for any $m,n \geqslant N$ we have $S - \epsilon < S_{NN} \leqslant S_{mn} \leqslant S_{m+n,m+n} \leqslant S.$

Finally, using a squeezing argument, we can show that the diagonal sum converges, $\sigma_n \to S,$ if and only if $S_{nn} \to S$. This follows because for $m > 2n$ we have $S_{nn} \leqslant \sigma_m \leqslant S_{mm}$ and $\sigma_n \leqslant S_{mm} \leqslant \sigma_{2m}.$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Sum along the values of $r=m+n$ As there are $r-1$ solutions to the equation $m+n=r$ with $m,n\ge 1$, we have: $$\sum_{n,m = 1}^{\infty} \frac{1}{(n+m)!}=\sum_{r= 2}^{\infty} \frac{r-1}{r!}=\sum_{r= 2}^{\infty}\biggl( \frac{1}{(r-1)!}-\frac{1}{r!}\biggr)$$ which is a telescoping series which simplifies to its initial term $1$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I've jjust seen I wrongly calculated the number of solution of $m+n=r$ (forgot that $m,n\ge 1$ , not $0$). I'll fix it in a moment. $\endgroup$ – Bernard Oct 28 '16 at 22:24
  • $\begingroup$ @Bernard: The second summation should be of $\frac{r-1}{r!}$, leading to $(e-1)-(e-2) = 1$. $\endgroup$ – Brian Tung Oct 28 '16 at 22:24
  • $\begingroup$ @Brian Tung: Yes, I forgot the constraints on $m$ and $n$. It's fixed now. Thanks to both for pointing it! $\endgroup$ – Bernard Oct 28 '16 at 22:30
  • $\begingroup$ Thank you but what I did is not very dissimilar than what you have done. My question is : is it good for a formal proof? $\endgroup$ – JohnK Oct 28 '16 at 22:32
  • $\begingroup$ I don't see how you do the initial step (1st equality) nor how you go from a double sum to a simple sum (2nd equality). $\endgroup$ – Bernard Oct 28 '16 at 22:41
1
$\begingroup$

I used the fact that the number of ways to sum with $k$ natural addends a number $n\in\Bbb N$ without restriction is $\binom{n-1}{k-1}$[1].

Then we have that $$\sum_{m,n\ge 1} \frac1{(m+n)!}=\sum_{k\ge 2}\frac{\binom{k-1}{2-1}}{k!}=\sum_{k\ge 2}\frac{k-1}{k!}$$

From here it follow the same solution.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ $k$ is always larger than 2 because of the restriction that both $m$ and $n$ are larger than $1$. $\endgroup$ – JohnK Oct 28 '16 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.