# Finding sum of infinite series by hand (non-geometric)

I'm reviewing for an exam currently and tried to find the convergence & sum of the series: $$\sum_{0}^{\infty} \frac{2^{n}}{n!}$$ I know from using the ratio test that the series does converge, but I'm at a loss as to how to find the sum. It's not geometric or telescoping, so I cannot use those associated formulas; I thought to try using the method in which you write out the first few terms and use them to find an expression for the partial sum of the series, but the original expression is so simple that the first 4 terms- $$S_{n} = \frac{1}{1}, \frac{2}{1}, \frac{4}{2}, \frac{8}{6}, \frac{16}{24}, ...$$ most easily simplify back into the original series' expression. I would really appreciate it if anyone can point me in the right direction, as I feel I must be missing the proper approach to this.

Many thanks in advance!

• No obvious answer. Use $e^x=\sum\limits_0^\infty \frac{x^n}{n!}$. Apr 2, 2023 at 22:08

Using only the series expansion for $$e$$ and the Cauchy product we write \begin{aligned} \sum_{n=0}^\infty\frac{2^n}{n!} &=\sum_{n=0}^\infty\frac{(1+1)^n}{n!}\\ &=\sum_{n=0}^\infty\sum_{k=0}^n\frac{1}{(n-k)!k!}\\ &=\left(\sum_{n=0}^\infty\frac{1}{n!}\right)\cdot\left(\sum_{k=0}^\infty\frac{1}{k!}\right)\\ &=(e)\cdot (e)\\ &=e^2. \end{aligned}
Generally, you'd want to think about treating a series like this (i.e., in the form $$f(x)=\sum a_n x^n$$) as a power series evaluated at a particular value of $$x$$; even if you don't immediately recognize the power series, this technique at least lets you try to solve for the function using its properties (under differentiation, say). Here (where $$a_n=1/n!$$), taking the derivative gives back the same power series, telling you that $$f'(x)=f(x)$$, or $$(\log f)'=1$$, or $$\log f = x + C$$, or $$f(x)=Ce^x$$ (not the same $$C$$). Together with $$f(0)=1$$, this tells you that in fact $$f(x)=e^x$$ and your sum evaluates to $$e^2 \approx 7.39$$.