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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?

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1 Answer 1

up vote 4 down vote accepted

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

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

Then

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

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

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

Let

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

and

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

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.

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The indices are correct, since the products apparently start at $1$ and not zero –  Cocopuffs Nov 17 '12 at 22:19
    
@Cocopuffs: Argh. I just got up, and clearly I’m not yet fully awake. –  Brian M. Scott Nov 17 '12 at 22:21
    
@brian-m-scott It happens - I think it's more likely there was a mistake in the problem given –  Cocopuffs Nov 17 '12 at 22:22
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@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. –  Brian M. Scott Nov 17 '12 at 22:55
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@Chistian: $$\sum_{n\ge 0}(-1)^n\frac{4^nx^{2n+1}}{(2n+1)!}=\sum_{n\ge 0}(-1)^n\frac{2^{(2n+1)-1}x^{2n+1}}{(2n+1)!}=\frac12\sum_{n\ge 0}(-1)^n\frac{(2x)^{2n+1}}{(2n+1)!}=\frac12\sin 2x$$ –  Brian M. Scott Nov 18 '12 at 0:34
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