$\sum u_n$ converge iff $\sum n(u_n-u_{n+1})$ converge Suppose that $(u_n)_{n \in \mathbb{N}}$ is a decreasing sequence and $lim_{n \to \infty} u_n = 0$. 
Prove that: $\sum u_n $ converges iff $\sum n(u_n-u_{n+1})$ converges.
The $\to$ way is simple to check, but I have trouble working on the other way.
 A: It suffices to show that $nu_n$ remains bounded. Fix $n$. Let $b_k = k(u_k - u_{k+1})$ and $B(t) = \sum_{k \leq t} b_k$. Partial summation gives
$$\begin{align*}
u_n &= u_{n+L+1} + \sum_{k=n}^{n+L} \frac{b_k}k \\
&= u_{n+L+1} + \frac{B(n+L)}{n+L} - \frac{B(n-1)}{n-1} + \int_{n-1}^{n+L} \frac{B(t)}{t^2} dt \\
&\leq u_{n+L+1} + \frac{B(n+L)}{n+L} + B(\infty) \int_{n-1}^\infty \frac{dt}{t^2} \\
&\leq u_{n+L+1} + \frac{B(\infty)}{n+L} + \frac{B(\infty)}{n-1}
\end{align*}$$
Now take $L$ sufficiently large so that $n u_{n+L+1} \leq 1$.
A: We invoke the following easy-to-prove theorem.

Tonelli's Theorem for Series. Let $a_{m,n} \geq 0$ for all $m, n \in \mathbb{N}$. Then
$$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n}
= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_{m,n} $$
regardless of whether they are finite or not.
Proof. Write $S_{M,N}=\sum_{m=1}^{M}\sum_{n=1}^{N} a_{m,n}$ and notice that
$$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n} = \sup_{M \geq 1}\sup_{N\geq 1} S_{M,N} \qquad\text{and}\qquad \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n} = \sup_{N\geq 1}\sup_{M \geq 1} S_{M,N}. $$
Then the desired conclusion follows by noting that supremums can be taken in arbitrary order without changing the value. $\square$

Using this, we find that whenever $u_n$ is non-negative and non-increasing,
\begin{align*}
\sum_{n=1}^{\infty} n (u_n - u_{n+1})
&= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \mathbf{1}_{\{ k \leq n\}} (u_n - u_{n+1}) \\
&= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \mathbf{1}_{\{ k \leq n\}} (u_n - u_{n+1}) \\
&= \sum_{k=1}^{\infty} (u_k - u_{\infty}),
\end{align*}
where $u_{\infty} := \lim_{n\to\infty} u_n$. So, if in addition that $u_{\infty} = 0$ holds, then this implies
$$ \sum_{n=1}^{\infty} n (u_n - u_{n+1}) = \sum_{k=1}^{\infty} u_k $$
regardless of the convergence of each side, and so, the desired claim follows.

For a more classical solution, check this:
Let $(u_n)_{n\geq 1}$ be a sequence of non-negative real numbers which is non-increasing and $u_n \to 0$ as $n\to\infty$. Then, As OP noted, summation by parts shows that
$$ \sum_{n=1}^{N} n (u_n - u_{n+1})
= - N u_{N+1} + \sum_{n=1}^{N} u_n
\leq \sum_{n=1}^{N} u_n, \tag{1}$$
and so, the convergence of $\sum_{n=1}^{\infty} u_n$ implies that of $\sum_{n=1}^{\infty} n(u_n - u_{n+1})$. Conversely, note that
\begin{align*}
\sum_{n=1}^{\infty} n (u_n - u_{n+1})
&= \left( \sum_{n=1}^{N} n (u_n - u_{n+1}) \right) + \left( \sum_{n=N+1}^{\infty} n (u_n - u_{n+1}) \right) \\
&\geq \left( - N u_{N+1} + \sum_{n=1}^{N} u_n \right) + \left( \sum_{n=N+1}^{\infty} N (u_n - u_{n+1}) \right) \\
&= \left( - N u_{N+1} + \sum_{n=1}^{N} u_n \right) + N u_{N+1} \\
&= \sum_{n=1}^{N} u_n,
\end{align*}
The second step follows from $\text{(1)}$ together with non-negativity of $u_n - u_{n+1}$, and the third step follows from the assumption that $u_n\to0$ as $n\to\infty$. Therefore this shows that the convergence of $\sum_{n=1}^{\infty} n(u_n - u_{n+1})$ implies that of $\sum_{n=1}^{\infty} u_n$.
A: Note that
$$
\begin{align}
\sum_{k=1}^nu_k-\sum_{k=1}^nk(u_k-u_{k+1})
&=\sum_{k=1}^n(ku_{k+1}-(k-1)u_k)\\
&=nu_{n+1}\tag1
\end{align}
$$
This answer shows that if $u_n$ is a decreasing sequence and $\sum\limits_{k=1}^nu_k$ converges, then $\lim\limits_{n\to\infty}nu_n=0$.
Suppose that $u_n$ is a decreasing sequence and $\sum\limits_{k=1}^nk(u_k-u_{k+1})$ converges, then
$$
\begin{align}
\lim_{n\to\infty}nu_n
&=\lim_{n\to\infty}n\sum_{k=n}^\infty(u_k-u_{k+1})\\
&\le\lim_{n\to\infty}\sum_{k=n}^\infty k(u_k-u_{k+1})\\[6pt]
&=0\tag2
\end{align}
$$
Thus, if either sum converges, $\lim\limits_{n\to\infty}nu_n=0$, and $(1)$ shows that the other sum also converges.
