Convergence of $\sum_{k=0}^{\infty}{x^k\over (k+1)!}$ How can I prove the convergence of $$\sum_{k=0}^{\infty}{x^k\over (k+1)!}$$ 
and what is the limit function? I think that I need to use the fact that $\sum_{k=0}^{\infty}{x^k\over k!}$ is a convergent series. 
Any comments, suggestions or hints would be really appreciated 
 A: $\left|\frac{x^k}{(k+1)!}\right|\leq \left|\frac{x^k}{(k)!}\right|$. Then use M-test and triangular inequality, namely, $\left|\sum\frac{x^k}{(k+1)!}\right|\leq \sum\left|\frac{x^k}{(k+1)!}\right|\leq \sum\left|\frac{x^k}{(k)!}\right|<\infty$ 
A: Hint: notice that $\sum\limits_{k=0}^{\infty}\frac{x^k}{(k+1)!}=\frac{1}{x}\sum\limits_{k=0}^{\infty}\frac{x^{k+1}}{(k+1)!}=\frac{1}{x}\sum\limits_{k=1}^{\infty}\frac{x^k}{k!}$.
A: So let's start with the formula for $e^u$:
$$\sum_{k=0}^\infty \frac{u^k}{k!}$$
The formula you provide is
$$\sum_{k=0}^\infty \frac{x^k}{(k+1)!}$$
$$=\frac{1}{x}\sum_{k=1}^\infty \frac{x^k}{k!}$$
$$=\frac{1}{x}\bigg(\sum_{k=0}^\infty \frac{x^k}{k!}-1\bigg)$$
$$=\frac{1}{x}\bigg(\sum_{k=0}^\infty \frac{x^k}{k!}-1\bigg)$$
$$=\frac{e^x - 1}{x}$$
$$***$$
As a sidenote, the following formula is a generalization of your's and is one of my favorites:
$$\sum_{k=0}^{\infty} \frac{x^k}{(k+n)!} = \frac{e^x x^{-n} (\Gamma(n)-\Gamma(n, x))}{\Gamma(n)}$$
Plugging in $n=1$,
$$\sum_{k=0}^{\infty} \frac{x^k}{(k+1)!} = \frac{e^x x^{-1} (\Gamma(1)-\Gamma(1, x))}{\Gamma(1)} = \frac{e^x(1-e^{-x})}{x}$$
$$ = \frac{e^x-e^{x-x}}{x} = \frac{e^x-1}{x}$$
