$\sum u_n$ 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: $$\sum _{k = 1}^{n} n(u_n-u_{n+1}) = -nu_{n+1} + \sum_{n=1}^{k} \leq \sum_{n=1}^{k} u_n,$$ since the term on the right side is bounded and the one on the left side is always positive, we have the convergence.

But I have trouble working on the other way. If we can prove $$\sum n(u_n-u_{n+1})$$ converge implies that $$\lim_{n\to\infty} nu_n = 0,$$ the other way will be proved.

• Can you show/sketch your proof for $\rightarrow$? It could help. Commented Sep 12, 2019 at 12:31
• $\sum _{k = 1}^{n} n(u_n-u_{n+1}) = -nu_{n+1} + \sum_{n=1}^{k} \leq \sum_{n=1}^{k} u_n$, since the term on the right side is bounded and the one on the left side is always positif, we have the convergence. Commented Sep 12, 2019 at 12:35
• If we can prove $\sum n(u_n-u_{n+1})$ converge implies that $lim_{n\to\infty} nu_n = 0$, the other way will be proved Commented Sep 12, 2019 at 12:42
• It's enough to show that convergence of $\sum n (u_n - u_{n + 1})$ implies $n u_{n +1}$ is bounded. Remember, we have a nonnegative sequence, so showing that \sum u_n is bounded implies convergence. Commented Sep 12, 2019 at 12:44

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.

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 $$n u_{n+L+1} \leq 1$$.

• I've come up with a proof: Let $v_n = n(u_n-u_{n_1})$, then $\frac{v_n}{n} \leq v_n$, so $\sum \frac{v_n}{n}$ converges, noting the sum by S. So we can write $u_n = \sum_{k=n}^{\infty} \frac{v_k}{k} \leq \frac{\sum_{k=n}^{\infty} v_k}{n} \leq \frac{S}{n}$, so $nu_n \leq S$. Commented Sep 12, 2019 at 13:11
• @pojoo You can write a separate answer with that. Commented Sep 12, 2019 at 15:34

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 $$\sum_{n=1}^{\infty} u_n$$.

Suppose $$a_n\searrow0$$ as $$n\rightarrow\infty$$. $$\sum_n a_n<\infty$$ iff $$na_n\xrightarrow{n\rightarrow\infty}0$$ and $$\sum_n n(a_n-a_{n+1})<\infty$$.
First, using Abel's summation by parts we have \begin{align} \sum^N_{k=1}a_k=\sum^N_{k=1}(k-(k-1))a_k=\sum^{N-1}_{k=1}k(a_k-a_{k+1}) + Na_N \tag{1}\label{one} \end{align}
To prove necessity, it is enough to show that $$na_n\xrightarrow{n\rightarrow\infty}0$$. Recall that from Cauchy's condensation theorem, convergence of $$\sum_na_n$$ and the fact that $$a_n$$ is nonincreasing is equivalent to convergence of $$\sum_k 2^k a_{2^k}$$. This in particular implies that $$2^ka_{2^k}\xrightarrow{k\rightarrow\infty}0$$. Now, for any $$n\in\mathbb{N}$$, there is a unique $$k_n\in\mathbb{Z}_+$$ such that $$2^{k_n}\leq n< 2^{k_n +1}$$; hence $$na_n\leq na_{2^{k_n}}\leq 2\,2^{k_n}a_{2^{k_n}}\xrightarrow{n\rightarrow\infty}0$$ The desired conclusion follows.