$\sum u_n$ converge iff $\sum n(u_n-u_{n+1})$ converge

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

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

Answer 4

Here is yet another answer.

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}

Sufficiency is then clear from \eqref{one}

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.