How can I prove the $\sum_{n=2}^{\infty }\frac{\log(n)^\frac{1}{k}}{n!}< (e-2)$ How can I prove the $$\sum_{n=2}^{\infty }\frac{\log(n)^\frac{1}{k}}{n!}< (e-2)$$ 
If $k>1$ 
 A: We may use Jensen's inequality. Since:
$$ f(x) = \exp\left(\frac{1}{k}\log\log x\right)=\left(\log x\right)^{\frac{1}{k}}\tag{1} $$
is a concave function over $x\geq 2$, we have:
$$ \frac{\lambda_2 f(2) + \lambda_3 f(3) + \ldots}{\lambda_2+\lambda_3+\ldots} \leq f\left(\frac{2 \lambda_2 + 3\lambda_3 +\ldots}{\lambda_2+\lambda_3+\ldots}\right) \tag{2}$$
or, by choosing $\lambda_m=\frac{1}{m!}$:
$$ \sum_{n\geq 2}\frac{(\log n)^{\frac{1}{k}}}{n!}\leq \left(\sum_{n\geq 2}\frac{1}{n!}\right)\cdot\log\left(\frac{\sum_{n\geq 2}\frac{n}{n!}}{\sum_{n\geq 2}\frac{1}{n!}}\right)^{\frac{1}{k}}\tag{3}$$
that is just:
$$ \sum_{n\geq 2}\frac{(\log n)^{\frac{1}{k}}}{n!}\leq (e-2)\cdot\left(\log\frac{e-1}{e-2}\right)^{\frac{1}{k}}. \tag{4}$$
This is stronger than needed, since $\log\frac{e-1}{e-2}\approx 0.872218 < 1$.
A: My first thought was to use Jensen's Inequality, as Jack did, but since he posted first, I looked for an alternate approach. 

Let
$$
f(\alpha)=\sum_{n=2}^\infty\frac{\log(n)^\alpha}{n!}\tag{1}
$$
First,
$$
f(0)=e-2\tag{2}
$$
Also, $f$ is convex because
$$
\begin{align}
f''(\alpha)
&=\sum_{n=2}^\infty\log(\log(n))^2\frac{\log(n)^\alpha}{n!}\\
&\gt0\tag{3}
\end{align}
$$
Using the fact that $n-1\ge\log(n)$, we can compute
$$
\begin{align}
f(1)
&=\sum_{n=2}^3\frac{\log(n)}{n!}+\sum_{n=4}^\infty\frac{\log(n)}{n!}\\
&\le\frac{\log(2)}2+\frac{\log(3)}6+\sum_{n=4}^\infty\frac{n-1}{n!}\\
&=\frac{\log(2)}2+\frac{\log(3)}6+\frac16\\[9pt]
&=0.696342305\\[12pt]
&\lt0.718281828\\[12pt]
&=e-2\tag{4}
\end{align}
$$
Since $f(0)=e-2$ and $f(1)\lt e-2$, then because $f$ is convex, we get that $f(\alpha)\lt e-2$ for all $0\lt\alpha\lt1$. This implies
$$
\bbox[5px,border:2px solid #C00000]{\sum_{n=2}^\infty\frac{\log(n)^{1/k}}{n!}\lt(e-2)}\tag{5}
$$
