Simplify the following series using the Cauchy product
- $$\sum\limits_{k=1}^\infty\frac{1}{k!}\cdot\sum\limits_{j=1}^\infty\frac{1}{j!}$$
- $$\sum\limits_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}\cdot\sum\limits_{j=0}^\infty\frac{(-1)^jx^{2j+1}}{(2j+1)!}$$
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?
Best Answer
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.