Find a counterexample to show the divergence of the following series This post is related to this one: prove the convergence of the following series.
As suggested by a user there, it's better to ask another question to see whether the series:
$$\sum\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^{\,p}}$$
(with $0<d_n<1$, $d_n$ being the general term of a divergent series) is divergent for at least one real $p \le 1$ and some $d_n$. It was shown there that the series converges for each real $p>1$. The original problem asks to show convergence for every $p > 0$. I was wondering if $p > 0$ was a typo for $p > 1$, so I started to study the case $p=1$ and hoped to prove that the series diverges, to settle the problem once for all for each $p > 0$ (a fortiori). It turns out that if we require $d_n$ to be monotone decreasing then the series converges for $p = 1$ as you can see in that post. That made me ask this question, to see if we can come up with a counterexample for the $0 < p \le 1$ case. Any help would be greatly appreciated!
 A: I find this problem highly interesting! Instead of suggesting a counterexample, I'll try to write up a proof for $p>0$.
Let $S_{n} = \sum_{j=0}^{n} d_{j}$. First, we observe
$$ (1+d_{j}) \ge 2^{d_{j}} $$
for all $d_{j}$. Indeed, $f(t) = 1+t$ and $g(t) = 2^{t}$ satisfy that $f(0) = g(0)$, $f(1) = g(1)$. Then the convexity of $g$ gives the desired property. Thus, we have
$$
[(1+d_{0}) \cdots (1+d_{n})]^{p} \ge 2^{p(d_{0} + \cdots + d_{n})}
$$
for all $n$. (Here $p > 0$ is used) Now we set a constant $K$ such that $$
2^{pt/2} \ge K'(t + 1)
$$
holds for all $t \ge 0$. For example, $K' = \min\{1, \frac{p}{2}\ln 2\}$ will do. This implies that $$
2^{p(d_{0} + \cdots + d_{n})} \ge K'^{2} (1 + d_{0} + \cdots + d_{n})^{2} \ge K'^{2} (d_{0} + \cdots + d_{n}) (d_{0} + \cdots + d_{n+1}) = K'^{2} S_{n} S_{n+1}.
$$
This implies that the series is dominated by $$
\frac{1}{K'^{2}}\sum_{n} \frac{S_{n+1} - S_{n}}{S_{n}S_{n+1}} = \frac{1}{K'^{2}}\sum_{n} \left(\frac{1}{S_{n}} - \frac{1}{S_{n+1}}\right).
$$
Since $S_{n} \rightarrow \infty$ as $n \rightarrow \infty$, telescoping will give us the desired result.
