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.
The intuition is already almost the proof.
- The cosine takes values between $-1$ and $1$, inclusive
- Taking an even power ensures that the values are between $0$ and $1$.
- Taking higher and higher powers will not affect those $x$ where the cosine squared is $1$, but all other values will converge to $0$
Thus $f(x):=\lim_{n\to\infty}\cos^{2n}(x)$ is a function that has value $0$ for all $x$, except that $f(x)=1$ if $\cos(x)=\pm1$, i.e. if $x$ is a multiple of $\pi$.
After this, we see that multiplying the argument with $\pi$ is done to have the value $1$ at integers instead of multiples of $\pi$. Then the multiplication with $m!$ serves the purpose to obtain $1$ for any integer multiple of $\frac1{m!}$. As $m\to \infty$, this exhausts the rationals (and only them).
Best Answer
Note that $$c_n=\sum_{k=0}^n a_kb_{n-k}$$ where $a_k$ is the coefficient of $x^k$ of the series of $\sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $\cos(x)$. Now if $n$ is even then
i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;
ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.
Hence $c_n=\sum_{k=0}^n 0=0$.
By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.