Help prove that $\lim_{n \to \infty} \sum_{k = 1}^{n/2} (\frac{ne}{k})^k(1-1.8\frac{\ln(n)}{n})^{k(n-k)} = 0$

Question

I am trying to prove that $$\lim_{n \to \infty} \sum_{k = 1}^{n/2} (\frac{ne}{k})^k(1-1.8\frac{\ln(n)}{n})^{k(n-k)} = 0,$$ but did not manage to get far. I tried using the same techniques as proposed in How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$? and $\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\ln k}{n} ( 1-\{\frac{n}{k} \} ) ( (1-\frac{k}{n} \{\frac{n}{k}\} )-\frac{1}{2})$, but did not see how these could be expanded to work in my case. Any hints in how to solve the limit are very welcome. I am especially struggling with the fact that the summation only goes to $n/2$, not to $n$.

Answer 1

The summands can be bounded as follows \begin{align*} & \le \left( {\frac{{ne}}{k}} \right)^k \exp \left( { - 1.8\frac{{\log n}}{n}k(n - k)} \right) = \left( {\frac{e}{k}} \right)^k \exp \left( {k\left( {1 - 1.8\frac{{n - k}}{n}} \right)\log n} \right). \end{align*} For $1 \le k \le \left[ {0.4n} \right]$ the right-hand side is at most $$ \left( {\frac{e}{k}} \right)^k \exp \left( { - \frac{1}{2}\log n} \right) = \left( {\frac{e}{k}} \right)^k \frac{1}{{\sqrt n }}, $$ whereas for $\left[ {0.4n} \right] < k \le \left[ {0.5n} \right]$, it is at most $$ \left( {\frac{e}{k}} \right)^k \exp \left( {\frac{n}{{20}}\log n} \right) \le \left( {\frac{e}{{0.4n}}} \right)^{0.4n} \exp \left( {\frac{n}{{20}}\log n} \right) = n^{ - 7n/20} \left( {\frac{e}{{0.4}}} \right)^{0.4n} < n^{ - 7n/20} 3^n . $$So your sum (which is positive) can be bounded from above by $$ \sum\limits_{k = 1}^{\left[ {0.4n} \right]} {\left( {\frac{e}{k}} \right)^k \frac{1}{{\sqrt n }}} + \sum\limits_{k = \left[ {0.4n} \right] + 1}^{\left[ {0.5n} \right]} {n^{ - 7n/20} 3^n } = \mathcal{O}(1)\frac{1}{{\sqrt n }} + \mathcal{O}(1)n^{1 - 7n/20} 3^n = \mathcal{O}(1)\frac{1}{{\sqrt n }}. $$