Convergence of series using domination Let $(x_n)_n$ be a sequence of $]0,+\infty[,y_n=\sum_{k=1}^nx_k$ such that $\lim_n y_n=+\infty.$
Let $p>1.$
Prove that $$\sum_{n}\dfrac{x_n}{y_n(\ln(y_n))^p}$$ converges.
Maybe the easiest way to prove it is to show that $\sum_n\dfrac{x_n}{y_n(\ln(y_n))^p}$ is dominated with a convergent series. Any ideas ?
 A: Take $q$ such that $y_{q-1} > 1$ and note that
$$S_m=\sum_{n=q}^m \frac{x_n}{y_n \ln^p y_n} = \sum_{n=q}^m \frac{y_n-y_{n-1}}{y_n \ln^p y_n} $$
Since $x \mapsto \frac{1}{x \ln^p x}$ is monotonically decreasing, we have
$$\frac{y_n-y_{n-1}}{y_n \ln^p y_n} \leqslant \int_{y_{n-1}}^{y_n}\frac{dx}{x \ln^px} = \frac{1}{p-1}\left(\frac{1}{\ln^{p-1} y_{n-1}}- \frac{1}{\ln^{p-1} y_n}\right),$$
and
$$S_m \leqslant \frac{1}{p-1}\left( \frac{1}{\ln^{p-1} y_{q-1}}- \frac{1}{\ln^{p-1} y_m}\right)$$
Since $y_m \to +\infty$ as $m \to \infty$, the sequence of partial sums converges with
$$\lim_{m \to \infty}S_m \leqslant \lim_{m \to \infty}\frac{1}{p-1}\left( \frac{1}{\ln^{p-1} y_{q-1}}- \frac{1}{\ln^{p-1} y_m}\right) = \frac{1}{(p-1)\ln^{p-1} y_{q-1}}$$
A: Consider the integer parts $[y_n]$ of all $y_n$. Assume they take values (excluding possible integer values $0, 1, 2$) $N_1<N_2<N_3<...$. Thus $N_1\geq 3$. Write the sum as
$$
\sum_{y_n<3}+\sum_{[y_n]=N_1}+\sum_{[y_n]=N_2}+\sum_{[y_n]=N_3}+...
$$
Ignore the first term. Start with
$$
\sum_{[y_n]=N_1}=\frac{x_{k_1}}{y_{k_1}(\log y_{k_1})^p}
+\frac{x_{k_1+1}}{y_{k_1+1}(\log y_{k_1+1})^p}+...+\frac{x_{k_1+r_1}}{y_{k_1+r_1}(\log y_{k_1+r_1})^p};
$$
since adding $x_{k_1+1}$, $x_{k_1+2}$, ... is not enough for the sum $y$ to jump from $N_1$ level to $N_1+1$ level, we see $x_{k_1+1}+...+x_{k_1+r_1}\leq 1$. So we estimate
$$
\sum_{[y_n]=N_1}\leq \frac{x_{k_1}+1}{y_{k_1}(\log y_{k_1})^p}.
$$
Thus the overall sum, ignoring the first sum, is bounded by
$$
\begin{aligned}
\frac{x_{k_1}+1}{y_{k_1}(\log y_{k_1})^p} & +\frac{x_{k_2}+1}{y_{k_2}(\log y_{k_2})^p}+
... \\
=& \frac{x_{k_1}}{y_{k_1}(\log y_{k_1})^p} +\frac{1}{y_{k_1}(\log y_{k_1})^p} 
+\frac{x_{k_2}}{y_{k_2}(\log y_{k_2})^p}+\frac{1}{y_{k_2}(\log y_{k_2})^p}+...\\
\leq & \frac{x_{k_1}}{N_1(\log N_1)^p} +\frac{1}{N_1(\log N_1)^p} 
+\frac{x_{k_2}}{N_2(\log N_2)^p}+\frac{1}{N_2(\log N_2)^p}+...\\
\end{aligned}
$$
The $\frac{1}{N_1(\log N_1)^p}+\frac{1}{N_2(\log N_2)^p}+...$
part is convergent. For the other terms, we can estimate for example
$$
\begin{aligned}
\frac{x_{k_2}}{N_2(\log N_2)^p}
<& \frac{N_2-N_1+2}{N_2(\log N_2)^p}= N_2-N_1+2 \text{ many } 
\frac{1}{N_2(\log N_2)^p}\\
<& \frac{2}{N_2(\log N_2)^p}+\frac{1}{(N_2-1)(\log (N_2-1))^p}
+...+\frac{1}{(N_1)(\log (N_1))^p};
\end{aligned}
$$
these terms fill the integer gap (if any) from $N_1$ to $N_2$.
So adding all together we estimate
$$
\frac{x_{k_1}}{N_1(\log N_1)^p}+
\frac{x_{k_2}}{N_1(\log N_2)^p}+...
<4\sum_{m=3}^\infty \frac{1}{m(\log m)^p}
$$
which is convergent as well.
