Convergence of the series $\sum_{n=1}^k\frac{k(n+1)}{k-n+1}\theta^{n}$

Question

I am trying to show that this inequality holds $$ \sum_{n=1}^k\frac{k(n+1)}{k-n+1}\theta^{n}<\frac{c}{(1-\theta)^2},\forall k>0,\theta\in(0,1) $$ where $c$ is some constant. I have done some numerical experiments on Matlab which suggests that $c$ may be not larger than 8. I also noticed that if we remove the $\frac{k}{k-n}$ term, then we have $\sum_{n=1}^k(n+1)\theta^{n}<\frac{1}{(1-\theta)^2}$. But I do not know how to proceed further. Any idea to show it? Thanks.

Answer 1

There's no such constant $c$. In fact, the corresponding $c_k$ grows like $\log k$ as $k\to\infty$, where $$f_k(\theta):=\sum_{n=1}^k\frac{(n+1)\theta^n}{k-n+1},\quad F_k(\theta):=k(1-\theta)^2 f_k(\theta),\quad c_k:=\max_{0\leqslant\theta\leqslant 1}F_k(\theta).$$ We prove the weaker $c_k=\Omega(\log k)$ by estimating $F_k(1-a/k)$, where $a>0$ is arbitrary.

Denote $g_k(\theta):=\sum_{n=1}^k\theta^{n-1}=(1-\theta^k)/(1-\theta)$ and $H_k:=\sum_{n=1}^k 1/n$. Then \begin{align} f_k(\theta)&=\sum_{n=1}^k\left(\frac{k+2}{k-n+1}-1\right)\theta^n \\&=(k+2)\sum_{n=1}^k\frac{\theta^{k-n+1}}{n}-\sum_{n=1}^k\theta^n \\&=(k+2)\theta^{k+1}\sum_{n=1}^k\frac{\theta^{-n}}{n}-\theta g_k(\theta). \end{align}

If $0<\theta<1$ then $\sum_{n=1}^k\theta^{-n}/n>H_k$, hence $f_k(\theta)>k H_k\theta^{k+1}-\theta g_k(\theta)$ and $$c_k\geqslant F_k(\theta)>\theta\big(k(1-\theta)\big)^2\left(H_k\theta^k-\frac{g_k(\theta)}{k}\right).$$

Now put $\theta=\theta_k:=1-a/k$; then $k(1-\theta_k)=a$ and, as $k\to\infty$, $${\theta_k}^k\to e^{-a},\quad\frac{g_k(\theta_k)}{k}\to\frac{1-e^{-a}}{a},\quad H_k\asymp\log k+\gamma+\ldots;$$ thus finally $\color{blue}{\liminf\limits_{k\to\infty} c_k/\log k}\geqslant a^2 e^{-a}\color{blue}{>0}$ as claimed (the maximum is at $a=2$).