If f(x) is an infinitely differentiable function at x=0 and $f^{(n)}(0)$ is the nth derivative of the function f at zero, then does the series below converge or diverge?
$\sum_{n=0}^{\infty} f^{(n+1)}(0)-f^{(n)}(0)$
For what kind of functions does the series converge? For what kind of functions that will make the above a divergent series, there is a known summation method?
Note that the sequence of the partial sums of the series $\sum\limits_{n=0}^\infty f^{(n+1)}(0)-f^{(n)}(0)$ is $$ s_n=\sum_{k=0}^n f^{(k+1)}(0)-f^{(k)}(0)=f^{(n+1)}(0)-f^{(0)}(0). $$ Thus summability of the series is equivalent to the convergence of the sequence $f^{(n)}(0)$.
Take for example $f(x)=\mathrm{e}^{ax}$, which is $C^\infty$ in $\mathbb R$. Then $$ f^{(n)}(0)=a^n. $$ This tends to $0$ is $|a|<1$, is equal to $1$ if $a=1$, does not tend anywhere if $a=-1$, tends to infinity if $a>1$.
In general, we can not say what the series does.