Let $\sum\limits_{k=1}^\infty a_n$ be a convergent series where $a_n\geq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $\sum\limits_{k=1}^\infty n(a_n-a_{n+1})$ also converges.

What I tried :

Let $(A_n)$ be the sequence of partial sums of the series $\sum\limits_{k=1}^\infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $\sum\limits_{k=1}^\infty n(a_n-a_{n+1})$.

Since $A_n=\sum\limits_{k=1}^n a_k$ and $B_n=\sum\limits_{k=1}^n k(a_k-a_{k+1})$ we get that:

\begin{align} B_n&=A_n-na_{n+1}\\ &=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\\ &>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\\ &=B_{n-1} \end{align}

we see that $(B_n)$ is a monotone increasing sequence $...(1)$


$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)

Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $\sum\limits_{k=1}^\infty n(a_n-a_{n+1})$ also converges

Is my proof correct?

  • $\begingroup$ Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} \to 0$ as $n\to\infty$. $\endgroup$
    – Winther
    Commented Dec 25, 2018 at 12:48
  • $\begingroup$ Thank you , and from the second one do we get that the sum of $\sum_{k=1}^\infty n(a_n-a_{n+1})$ equals the sum of $\sum_{k=1}^\infty a_n$? $\endgroup$ Commented Dec 25, 2018 at 13:00
  • 1
    $\begingroup$ Taking the limit $n\to\infty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} \to 0$ is to show that $\sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts. $\endgroup$
    – Winther
    Commented Dec 25, 2018 at 13:02
  • $\begingroup$ I can use another method to prove that $na_n\to0$ when $n \to \infty$ without using the convergence of the series $\sum_{k=1}^\infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n \to \infty $ implies that also the sequnce $(B_n)$ converges right? $\endgroup$ Commented Dec 25, 2018 at 13:18
  • $\begingroup$ Yes that's right $\endgroup$
    – Winther
    Commented Dec 25, 2018 at 13:19

2 Answers 2


Some comments:

  1. The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$\ldots$") and simplifying.

  2. Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n \le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).

  3. You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.

  4. You could also further note that, if $\lim B_n < \lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).

  • 1
    $\begingroup$ "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence? $\endgroup$ Commented Dec 25, 2018 at 13:12
  • 1
    $\begingroup$ @MathsSurvivor If $A_n \to A$ and $B_n \to B$ with $A \neq B$, then $$\frac{a_{n+1}}{\frac{1}{n}} = na_{n+1} = A_n - B_n \to A - B \neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging. $\endgroup$ Commented Dec 25, 2018 at 14:02

Note that we can use $n=\sum_{k=1}^n (1)$ to write

$$\begin{align} \sum_{n=1}^N n(a_{n+1}-a_n)&=\sum_{n=1}^N \sum_{k=1}^n(a_{n+1}-a_n)\\\\ &=\sum_{k=1}^N \sum_{n=k}^N (a_{n+1}-a_n)\\\\ &=\sum_{k=1}^N (a_{N+1}-a_k)\\\\ &=Na_{N+1}-\sum_{k=1}^N a_k\tag1 \end{align}$$

Inasmuch as $a_n\ge 0$ monotonically decreases to $0$, and $\sum_{k=1}^n a_n<\infty$, we have $\lim_{n\to\infty }na_n=0$. Hence, using $(1)$, we see that

$$\begin{align} \lim_{N\to \infty }\sum_{n=1}^N n(a_{n+1}-a_n)&=\lim_{N\to\infty}\left(Na_{N+1}-\sum_{k=1}^N a_k\right)\\\\ &=-\sum_{n=1}^\infty a_n \end{align}$$

from which we conclude that $\sum_{n=1}^\infty n(a_{n+1}-a_n)$ converges.

  • $\begingroup$ Winther's comment on the question suggests that showing $\lim n a_n = 0$ is often done by first establishing $\sum n(a_n - a_{n+1}) = \sum a_n$. Out of curiosity, how would you establish $\lim n a_n = 0$? $\endgroup$ Commented Dec 25, 2018 at 22:44
  • $\begingroup$ Note that $$(2n)a_{2n}\le 2\sum_{n+1}^{2n}a_k\to 0$$since $a_n\ge0$ is monotonic and $\sum_n a_n<\infty.$. $\endgroup$
    – Mark Viola
    Commented Dec 26, 2018 at 4:05

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