Monotonicity of a integral function $p_{k}$ For $k \ge 2$, we have $p_{k}$ in terms of
$$
p_{k} = k \exp \left(-2 \sum_{i = 1}^{k-1} \frac{1}{i}\right) \int_{0}^{1} \exp\left(2 \sum_{i=1}^{k-1} \frac{s^i}{i}\right)
$$
By computing $p_k$ term by term, we can have $p_2 = 0.865$, $p_3 = 0.824$, $p_4 = 0.804$ ... $p_{10} = 0.770$, $p_{100} = 0.748$. It's clear that the sequence is monotone decreasing when $k$ goes to infinity, but $p_k$ is somehow hard to analyze.
$p_k$ is actually the solution of an ordinary differential equation
$$
p'(t) = \frac{\left(k - 1\right) - \left(k - 2 \right) t + 2 t^k}{t \left(1 - t \right)} p \left(t \right) + \frac{k t^{k - 1}}{ \left(1 - t \right)^3}
$$
 A: Yes, it's quite hard. Note that we can rewrite it as $$p_{k}=k\exp\left(-2H_{k-1}\right)\int_{0}^{1}\exp\left(2\sum_{i=1}^{k-1}\frac{s^{i}}{i}\right)ds
 $$ where $H_{n}
 $ is the $n
 $-th harmonic number and using the well-known bound $$H_{n}=\log\left(n\right)+O\left(1\right)
 $$ we get, as $k\longrightarrow\infty
 $, $$p_{k}=\frac{k}{\left(k-1\right)^{2}}e^{O\left(1\right)}\int_{0}^{1}\exp\left(2\sum_{i=1}^{k-1}\frac{s^{i}}{i}\right)ds.
 $$ The problem is the estimation of the integral. Note that, if $k
 $ is big, then $$\sum_{i=1}^{k-1}\frac{s^{i}}{i}\approx-\log\left(1-s\right)
 $$ and so $$\int_{0}^{1}\exp\left(2\sum_{i=1}^{k-1}\frac{s^{i}}{i}\right)ds\approx\int_{0}^{1}\frac{1}{\left(1-s\right)^{2}}ds
 $$ and the last integral doesn't converge. Using the closed form $$2\sum_{i=1}^{k-1}\frac{s^{i}}{i}=-2\log\left(1-s\right)-2s^{k}\Phi\left(s,1,k\right)
 $$ where $ \Phi\left(a,b,c\right)
 $ is the Lerch trascendent, for a precise control of the integral we have to study $$\int_{0}^{1}\exp\left(2\sum_{i=1}^{k-1}\frac{s^{i}}{i}\right)ds=\int_{0}^{1}\frac{1}{\left(1-s\right)^{2}}\exp\left(-2s^{k}\Phi\left(s,1,k\right)\right)ds
 $$ and I think it's not trivial.
