Task from an old exam:
Calculate (express without an infinite sum): $$\frac{1}{2! \cdot 2} +
\frac{1}{4! \cdot 4} + \frac{1}{6! \cdot 8} + \frac{1}{8! \cdot 16} +
…$$
I think this means on the way to the solution, we are allowed to use the sum symbol but the final result may not be in a sum symbol. Else I don't see another way of solving this task.
In sum, the thing above would be:
$$\sum_{n=1}^{\infty} \frac{1}{(2n)!\cdot 2^{n}}$$
And now we need to do something so the sum symbol is eliminated but I have no idea what it could be…
What about seeing $$\frac{1}{(2n)!\cdot 2^{n}}$$ as a function and using taylor series on this? But no it would be too hard to derivate something like that and with factorial.. and I don't think it makes sense saying "Hey, let's replace this sum symbol with a function and say this thing is a function now!"
But what else can I do in a situation like this? If you are doing something complicated, please do explain me. I have big troubles in understanding things.
Best Answer
Note that the sum is the even terms of $f(x)=\sum \frac{x^n}{n!}$ evaluated at $x=\sqrt{2}/2$ (but ignoring the $0$th term). This is the Taylor series for $e^x$. To eliminate the odd terms, we take $(f(x)+f(-x))/2=\frac{e^x+e^{-x}}{2}$. To remove the $0$th term, we subtract $f(0)$. Putting everything together, the sum is $\frac{e^{\sqrt 2/2}+e^{-\sqrt 2/2}}{2}-1$.