Bound on $\sum_{k = 2}^n (en)^k k^{-Cn/\log{n} - k - 1/2}$

Question

As the title suggests, I'm trying to establish a good bound on

\begin{equation} S(n) = \sum_{k = 2}^n (en)^k k^{-Cn/\log{n} - k - 1/2}, \end{equation}

where $C$ is some reasonably large positive constant. In particular I would like to have $S(n) = o(1)$, i.e.,

\begin{equation} \lim_{n \to \infty} S(n) = 0, \end{equation}

which numerical evaluations suggest to be the case.

Moreover it appears to be the case that the terms in the series are monotonically decreasing; if this were true my claim follows trivially (replace all terms by the one at $k = 2$ and check) but verifying the ''monotone decrease conjecture'' is again a difficult task in itself.

I appreciate any ideas on how to tackle this problem.

EDIT: The sequence terms are unfortunately not monotonically decreasing as can be verified by studying the logarithm of the latter but the conjecture about the term at $k = 2$ being largest still stands.

Answer 1

In spite of my previous comment, I do think your intuition is correct.

Attempted proof (sorry but it is very ugly): Again, let $r(d)=\frac{s_2}{s_{2+d}}$. As you noted, $$ r(d)=\left(\frac{2+d}{en}\right)^d\left(1+\frac{d}{2}\right)^{Cn/\log n + 2 + 1/2} $$

Taking logarithm, we get $$ \log r(d)=d\left(\log\left(\frac{2+d}{en}\right)\right)+ (Cn/\log n+2+1/2)\left(\log\left(\frac{2+d}{2}\right)\right) $$ where $d\in [0,n-2]$. Now, let $\lambda=\frac{2+d}{n}\in[2/n,1]$, or, equivalently, $d=\lambda n - 2$. Thus, $$ \log r(d)=(\lambda n - 2)\left(\log\left(\frac{\lambda}{e}\right)\right) + (Cn/\log n+2+1/2)\left(\log\left(\frac{\lambda n}{2}\right)\right):=f(\lambda) $$

Cleaning this mess a bit, we get $$ f(\lambda)=(\lambda n -2)(\log\lambda - 1) + (Cn/\log n+2+1/2)(\log\lambda + \log n -\log2)\\ = \lambda n(\log\lambda - 1) -2\log\lambda+2+(Cn/\log n + 2 + 1/2)\log\lambda\\ + Cn + 2.5\log n-2.5\log 2\\ =n(\lambda \log\lambda - \lambda) + (Cn/\log n + 0.5)\log\lambda + D(n) $$ where $D(n)$ depends only on $n$.

Taking derivative, we get $f'(\lambda)=n\log\lambda + \frac{1}{\lambda}(Cn/\log n + 0.5)$, and the second derivative is $f''(\lambda)=\frac{n}{\lambda} - \frac{1}{\lambda^2}(Cn/\log n + 0.5)=\frac{n}{\lambda^2}(\lambda-C/\log n + 0.5/n)$

Hence. If $n$ is large enough, $f''(\lambda)>0$ for all $\lambda \ge 2/n$. Ie, $f'$ is increasing. Now, for such $n$, $$ f'\left(\frac{2}{n}\right)=n\log 2 - n\log n + \frac{n}{2}(Cn/\log n + 0.5)\\ =n\log 2 - n\log n + \frac{Cn^2}{2\log n} + n$$ which is bigger than $0$ for large enough $n$. And since $f'$ increasing, $f'>0$ in $[2/n,1]$. Ie, $f$ is increasing. Ie, for large enough $n$, $f(\lambda)\ge 0$ for $\lambda \ge 2/n$. Hence $r(d)\ge 1$ for $d\in [0,n-2]$. Hence $S(n)=o(1)$.

Answer 2

This seems to do it:

To argue that the second term $s_2$ is indeed largest consider the ratio

\begin{equation} r(d) = \frac{s_2}{s_{2 + d}} \end{equation}

for some $d \leq n - 2$. Our goal is to establish $r(d) \geq 1$ for $d \geq 1$. We ignore the $-1/2$ contribution in the exponent and find that

\begin{equation} r(d) = \left(1 + \frac{d}{2}\right)^{Cn/\log{n} + 2} \left(\frac{2 + d}{en}\right)^d. \end{equation}

Observe that

\begin{equation} \left(\frac{2 + d}{en}\right)^d \end{equation}

is increasing in $d$, so we define

\begin{equation} \tilde{r}(d) = \left(1 + \frac{d}{2}\right)^{Cn/\log{n}} \frac{3}{en} \end{equation}

and seek to satisfy the stricter requirement $\tilde{r}(d) \geq 1$. Let $f(n)$ denote the first value for which the inequality holds true. We find that

\begin{equation} f(n) = 2\left(\left(\frac{en}{3}\right)^{\frac{\log{n}}{Cn}} - 1\right). \end{equation}

Since

\begin{equation} \lim_{n \to \infty} \left(\frac{en}{3}\right)^{\frac{\log{n}}{Cn}} = 1 \end{equation}

we establish the existance of some $n_0$ s.t. $\tilde{r}(d) \geq 1$ for $n \geq n_0$ and hence we indeed find that $r(d) \geq 1$ for large $n$. The final claim ($\lim_{n \to \infty} S_n = 0$) can then be recovered as outlined in the question.