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How can I prove the $$\sum_{n=2}^{\infty }\frac{\log(n)^\frac{1}{k}}{n!}< (e-2)$$ If $k>1$

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  • $\begingroup$ Just a quick stab gives me: For $n \ge 2$ we know that $\log(n)^{1/k}/n < 1$. Thus $$\sum_{n=2}^\infty \frac{\log(n)^{1/k}}{n} \frac{1}{(n-1)!} \le \sum_{n=2}^\infty \frac{1}{(n-1)!} = e-1$$ $\endgroup$ – Joel Feb 23 '15 at 15:44
  • $\begingroup$ @SamiBenRomdhane I checked it but I couldn't understand why is not correct $\endgroup$ – user189855 Feb 23 '15 at 15:46
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    $\begingroup$ Numerical calculations show that this inequality is indeed correct. $\endgroup$ – sranthrop Feb 23 '15 at 15:47
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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 $\log\frac{e-1}{e-2}\approx 0.872218 < 1$.

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    $\begingroup$ Very nice and tighter! +1 $\endgroup$ – Macavity Feb 23 '15 at 15:55
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    $\begingroup$ Not that much stronger as $k$ gets bigger, but strong enough to get strict inequality. (+1) $\endgroup$ – robjohn Feb 23 '15 at 16:58
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    $\begingroup$ @Jack D'Aurizio Bravo ! +1. $\endgroup$ – Olivier Oloa Feb 23 '15 at 17:03
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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 $$ \bbox[5px,border:2px solid #C00000]{\sum_{n=2}^\infty\frac{\log(n)^{1/k}}{n!}\lt(e-2)}\tag{5} $$

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