Asymptotic behavior of $\sum_{k=1}^{n^2}\frac{q^{n^2-k}(1-q^k)}{k(1-q^{n^2})}$ when $q=1-\frac{\log(n)}{n}$

Question

Let $q=1-\dfrac{\log(n)}{n}$.

Numerical simulations indicate that that

\begin{align*} \lim_{n \to \infty}\displaystyle\sum_{k=1}^{n^2}\dfrac{q^{n^2-k}(1-q^k)}{k(1-q^{n^2})} = 0 \end{align*} in a monotone decreasing manner. But I wonder about the rate of convergence. Any ideas?

Answer 1

Your sum is basically $(1-q^{n^2})\sum_{k=0}^{n^2-1}{\frac{q^{k}}{n^2-k}}-\frac{q^{n^2}H_{n^2}}{1-q^{n^2}}$.

But $q^{n^2} \leq e^{-n^2 \cdot ((\ln{n})/n)}=n^{-n}$, so the second term is $O(n^{-n}\ln{n})$.

For the first term, the same computation shows that $q^{-3n} \leq n^{-3}$, so the sum is less than $\sum_{k=3n}^{n^2-1}{q^k}+(1+O(n^{-n}))\sum_{k=0}^{3n-1}{\frac{q^k}{n^2-k}}+O(n^{-n}\ln{n})$. The first sum is thus at most $O((n^2\ln{n})^{-1})$. The second sum is at most $(n^2-3n)^{-1}\cdot \frac{n}{\ln{n}}$, and therefore the total is $O((n\ln{n})^{-1})$.

To make the estimate more precise, we see that the dominant term is the sum $\sum_{k=0}^{3n-1}{\frac{q^k}{n^2-k}}$.

Note that $\sum_{k=0}^{3n-1}{\frac{1}{n^2-k}-\frac{1}{n^2}} \leq \frac{1}{n^3(n-3)}\frac{3n(3n-1)}{2} = O(n^{-2})$, so that the first sum is $O(n^{-2})+n^{-2}(1+O(n^n))\sum_{k=0}^{3n-1}{q^k}=O(n^{-2})+\frac{n}{\ln{n}}(1-q^{3n})=\frac{1}{n\ln{n}}+O(n^{-2})$.

So your sequence is equivalent to $\frac{1}{n\ln{n}}$.

Answer 2

This is just from numerical simulation.

Looking at the partial sums $S_n$, it seems that $$\log(S_n)=a-b\,n^c$$ could be quite good $(R^2=0.999994)$. $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 3.24689 & 0.13016 & \{2.98264,3.51113\} \\ b & 1.59206 & 0.01702 & \{1.55752,1.62660\} \\ c & 1.24158 & 0.00279 & \{1.23590,1.24725\} \\ \end{array}$$ For sure, the exponent $c$ is hidding some logarithms.