Recently, I tried to evaluate a limit proposed by MSE user Black Emperor. In the process of evaluating the limit, I have obtained the following equality.
$$
\lim_{N\rightarrow \infty} \sum_{n=0}^{N-2}{\frac{\left( -1 \right) ^n}{\left( 2n+1 \right) !}\zeta \left( N-n \right)}=\sin\left(1\right)
$$
But I believe this can applied to a broader case of limits.
Namely, if we have a function $f$ that is holomorphic around zero with a infinite radius of convergence. The following should hold
\begin{align*}
\lim_{N\rightarrow \infty} \sum_{n=0}^{N-2}{\frac{1}{n!}\zeta \left( N-n \right)}f^{\left( n \right)}\left( 0 \right)
=&\lim_{N\rightarrow \infty} \sum_{n=0}^{N-2}{\frac{1}{n!}\sum_{k=1}^{\infty}{\frac{k^n}{k^N}}}f^{\left( n \right)}\left( 0 \right)
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^N}\sum_{n=0}^{N-2}{\frac{1}{n!}f^{\left( n \right)}\left( 0 \right) k^n}}
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^N}\sum_{n=0}^{\infty}{\frac{1}{n!}f^{\left( n \right)}\left( 0 \right) k^n}}
\\
=&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{f\left( k \right)}{k^N}}=f\left( 1 \right)
\end{align*}
I have verified this result numerically for a some functions, and it seems that this equality holds for functions like:
$\mathbb{sin}\left(x\right)$, $\mathbb{cos}\left(x\right)$, $\mathbb{exp}\left(x\right)$, $\mathbb{sinh}\left(x\right)$, $\mathbb{cosh}\left(x\right)$, $\sqrt{x+1}$, $\mathbb{arctan}\left(x\right)$, $\mathbb{ln}\left(x\right)$, $\mathbb{erf}\left(x\right)$.
However, it did no longer work when I plugged $\mathrm{W}\left(x\right)$ in the Lambert W function.
I have three question here.
- Is this correct?
- As you can see, much of the examples do not have a infinite radius of convergence, yet, the equality still holds. Therefore, What's the ture criteria of this equality?
- Can we generalize this equality to a bigger family of functions?
You can use Tannery's theorem to get a sufficient condition. Your sum is $$ \sum\limits_{n = 0}^{N - 2} {\zeta (N - n)\frac{{f^{(n)} (0)}}{{n!}}} = \sum\limits_{n = 0}^\infty {\chi _{\left[ {0 \le n \le N - 2} \right]} \zeta (N - n)\frac{{f^{(n)} (0)}}{{n!}}} $$ and we have $$ \left| {\chi _{\left[ {0 \le n \le N - 2} \right]} \zeta (N - n)\frac{{f^{(n)} (0)}}{{n!}}} \right| \le 2\frac{{|f^{(n)} (0)|}}{{n!}}. $$ Thus, the absolute convergence of the Maclaurin series of $f(x)$ at $x=1$ is enough.