Simplify the following series using the Cauchy product

  1. $$\sum\limits_{k=1}^\infty\frac{1}{k!}\cdot\sum\limits_{j=1}^\infty\frac{1}{j!}$$
  2. $$\sum\limits_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}\cdot\sum\limits_{j=0}^\infty\frac{(-1)^jx^{2j+1}}{(2j+1)!}$$

Which relations did you compute with these two series?

Problem 1

First of all we define $$a_k=\frac{1}{k!}\quad\text{and}\quad b_j=\frac{1}{j!}$$ and now we have the series $$\left(\sum\limits_{n=0}^\infty c_{n}\right)=\left(\sum\limits_{k=0}^\infty a_{k+1}\right)\left(\sum\limits_{j=0}^\infty b_{j+1}\right)$$ where I think that $$c_n=\sum\limits_{m=0}^na_{n-m+1}b_{m+1}=\sum\limits_{m=0}^n\frac{1}{(n-m+1)!}\cdot\frac{1}{(m+1)!}=\sum\limits_{m=0}^n\frac{\binom{n}{m+1}}{n!}$$

however I thought that it will be something like $\exp(1)\cdot\exp(1)=\exp(2)$ but the indexes $k$ and $j$ start at $1$ and not $0$ so I am really confused how to simplify my solution (is it even correct?) and what the result will be.

I would like to hear some hints what to do here.

Problem 2

Knowing from Wolframalpha that the series equals to $\sin(x)\cos(x)=1/2\sin(2x)$ I computed this:

$$\begin{align} \sum\limits_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}\cdot\sum\limits_{j=0}^\infty\frac{(-1)^jx^{2j+1}}{(2j+1)!} &= \sum\limits_{n=0}^\infty\sum\limits_{k=0}^n(-1)^k\frac{x^{2k}}{(2k)!}(-1)^{n-k}\frac{x^{2(n-k)+1}}{(2(n-k)+1)!} \\ &= \sum\limits_{n=0}^\infty(-1)^n\frac{1}{(2n+1)!}\sum\limits_{k=0}^n\binom{2n+1}{2k}x^{2k}x^{2(n-k)+1} \\ &= \sum\limits_{n=0}^\infty(-1)^n\frac{4^nx^{2n+1}}{(2n+1)!} \end{align}$$

How should I do the last steps?


The problem is that the indices start at $1$ instead of $0$. I find it easiest to work with the sums starting at $k=0$ and $j=0$ and then adjust the result.

Setting $c_n=\sum_{i=0}^na_{n-i}b_i$, we have

$$\begin{align*} \left(\sum_{k\ge 0}a_k\right)\left(\sum_{j\ge 0}b_j\right)&=\sum_{n\ge 0}c_n\\ &=\sum_{n\ge 0}\sum_{i=0}^n\left(\frac1{(n-i)!}\cdot\frac1{i!}\right)\\ &=\sum_{n\ge 0}\sum_{i=0}^n\frac1{i!(n-i)!}\\ &=\sum_{n\ge 0}\sum_{i=0}^n\binom{n}i\frac1{n!}\\ &=\sum_{n\ge 0}\frac1{n!}\sum_{i=0}^n\binom{n}i\\ &=\sum_{n\ge 0}\frac{2^n}{n!}\\ &=e^2\;, \end{align*}$$


$$\begin{align*} \left(\sum_{k\ge 1}a_k\right)\left(\sum_{j\ge 1}b_j\right)&=\left(\sum_{k\ge 0}a_k-1\right)\left(\sum_{j\ge 0}b_j-1\right)\\ &=\sum_{n\ge 0}\frac{2^n}{n!}-\sum_{k\ge 0}a_k-\sum_{j\ge 0}b_j+1\\ &=\sum_{n\ge 0}\frac{2^n}{n!}-2\sum_{k\ge 0}\frac1{k!}+1\\ &=e^2-2e+1\;, \end{align*}$$

just as it should, since $\sum_{k\ge 1}\frac1{k!}=e-1$.

Added: Here’s a start on the second problem.


$$a_k=\begin{cases} \frac{(-1)^{k/2}}{k!},&\text{if }k\text{ is even}\\ 0,&\text{otherwise} \end{cases}$$


$$b_j=\begin{cases} \frac{(-1)^{(j-1)/2}}{j!},&\text{if }k\text{ is odd}\\ 0,&\text{otherwise}\;, \end{cases}$$

so that you’re looking at the product

$$\left(\sum_{k\ge 0}a_kx^k\right)\left(\sum_{j\ge 0}b_jx^j\right)\;.$$

The coefficient of $x^n$ in that product is $$c_n=\sum_{i=0}^na_{n-i}b_i\;.$$

When $n$ is even, $n-i$ and $i$ have the same parity, so $a_{n-i}b_i=0$. If $n$ is odd, $n-i$ and $i$ have opposite parity; when $i$ is even, $a_{n-i}b_i=0$, but when $i$ is odd, you get non-zero terms.

  • $\begingroup$ The indices are correct, since the products apparently start at $1$ and not zero $\endgroup$ – Cocopuffs Nov 17 '12 at 22:19
  • $\begingroup$ @Cocopuffs: Argh. I just got up, and clearly I’m not yet fully awake. $\endgroup$ – Brian M. Scott Nov 17 '12 at 22:21
  • $\begingroup$ @brian-m-scott It happens - I think it's more likely there was a mistake in the problem given $\endgroup$ – Cocopuffs Nov 17 '12 at 22:22
  • 1
    $\begingroup$ @Chistian: Sorry: that’s a typo; I must have changed $e$ to $e-1$ in too many places when I edited last time, but it’s fixed now. $\endgroup$ – Brian M. Scott Nov 17 '12 at 22:55
  • 1
    $\begingroup$ @Chistian: $$\sum_{n\ge 0}(-1)^n\frac{4^nx^{2n+1}}{(2n+1)!}=\sum_{n\ge 0}(-1)^n\frac{2^{(2n+1)-1}x^{2n+1}}{(2n+1)!}=\frac12\sum_{n\ge 0}(-1)^n\frac{(2x)^{2n+1}}{(2n+1)!}=\frac12\sin 2x$$ $\endgroup$ – Brian M. Scott Nov 18 '12 at 0:34

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.