prove the convergence of the following series If $0 < d_n < 1$ with $\sum d_n$ divergent, then the two series $$\sum d_{n+1}\left[(1-d_0)\cdots(1-d_n)\right]^p$$
$$\sum\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p}$$
are convergent, for every $p>0$. I think it's enough to show the convergence of the first one, because by substituting $1-d_n' = \frac{1}{1+d_n}$ one can reduce the second series to the first one (almost, except for a bounded factor). I don't really have any ideas on how to prove this. Can you help me?
 A: Let $S_0 = d_0$ and $S_n = \sum_{j=0}^n d_j $.
We have $(1+d_0)\cdots(1+d_n)\geqslant 1 + d_0 + \cdots + d_n \geqslant d_0 + \cdots + d_n + d_{n+1} = S_{n+1}$.
When $p > 1$ there exists a positive integer $m$ such that $\frac{1}{m} < p-1$ and
$$\sum_{n=0}^N\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant \sum_{n=0}^N\frac{S_{n+1}- S_n}{S_{n+1} ^p}  \leqslant  \sum_{n=0}^N\frac{S_{n+1}- S_n}{S_{n+1}S_n ^{p-1}} \\\leqslant  \sum_{n=0}^N\frac{S_{n+1}- S_n}{S_{n+1}S_n ^{1/m}}$$
Note that
$$\frac{S_{n+1}- S_n}{S_{n+1}S_n ^{1/m}}= \frac{1- \frac{S_{n}}{S_{n+1}}}{S_n^{1/m}}= \frac{1- \frac{S_{n}}{S_{n+1}}}{1- \frac{S_{n}^{1/m}}{S_{n+1}^{1/m}}}\left(\frac{1}{S_{n}^{1/m}} - \frac{1}{S_{n+1}^{1/m}} \right), $$
where the term $x = \frac{S_n^{1/m}}{S_{n+1}^{1/m}}= \left(\frac{S_n}{S_{n+1}}\right)^{1/m} \in (0,1) $ since the sequence $S_n$ is increasing .
By Bernoullis' inequality, we have $x^m = [1- (1-x)]^m \geqslant 1 - m(1-x)$ which implies that $1- x^m \leqslant m(1-x)$ and, substituting for $x$,
$$1- \frac{S_{n}}{S_{n+1}} \leqslant m\left(1- \frac{S_{n}^{1/m}}{S_{n+1}^{1/m}}\right)$$
Thus,
$$\sum_{n=0}^N\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant m\sum_{n=0}^N\left(\frac{1}{S_{n}^{1/m}} - \frac{1}{S_{n+1}^{1/m}} \right) = \frac{m}{d_0^{1/m}} - \frac{m}{S_{N+1}^{1/m}}$$
The series $\sum d_n$ diverges to $+\infty$ which implies that $m/S_{N+1}^{1/m} \to 0$ as $N \to \infty$ and the sum on the LHS converges, with
$$\sum_{n=0}^\infty\frac{d_{n+1}}{\left[(1+d_0)\cdots(1+d_n)\right]^p} \leqslant \frac{m}{d_0^{1/m}}$$

For the first series, taking $d_n' = d_n/(1-d_n)$, we have
$$1- d_n = \frac{1}{1+d_n'}, \quad d_{n+1} = \frac{d_{n+1}'}{1+d_{n+1}'},$$
and
$$d_{n+1}[(1- d_0) \cdots (1-d_n)]^p = \frac{d_{n+1}'}{\left[(1+d_0')\cdots(1+d_n')\right]^p(1+d_{n+1}')} \leqslant  \frac{d_{n+1}'}{\left[(1+d_0')\cdots(1+d_n')\right]^p}$$
Now you can prove convergence using the result for the second series, after showing that divergence of $\sum d_n$ implies divergence of $\sum d_n'$.
A: I'm going to show that we cannot prove in general that the series diverges for $p = 1$ as @RRL suspected. We can prove that the series $$\sum_{n=1}^{\infty}\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}$$ is for $a_n$ positive invariably convergent. One can show easily by induction that, if we let $S_n$ be the partial sums of the series, then $S_n = 1-\frac1{D_n}$, where $D_n = (1+a_1)\cdots(1+a_n)$. Thus, since the product $\prod(1+a_i)$ either diverges to $+\infty$ or converges, it follows that the sum converges. We can use this result for the problem. Let $d_n$ be the general term of a divergent series; if $d_n$ is monotone decreasing, then $$\sum \frac{d_{n+1}}{(1+d_0)(1+d_1)\cdots(1+d_n)}\le \frac1{(1+d_0)}\sum \frac{d_{n}}{(1+d_1)\cdots(1+d_n)}$$ and the result follows by the previous.
I will edit this post if I can come up with a proof for $p \le 1$.
