# Let $\sum_{k=1}^\infty a_n$ be convergent show that $\sum_{k=1}^\infty n(a_n-a_{n+1})$ converges

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 $$...(1)$$

and

$$B_n=A_n-na_{n+1} this implies that the sequence $$(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?

• 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 $n\to\infty$. – Winther Dec 25 '18 at 12:48
• 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 $\sum_{k=1}^\infty a_n$? – Maths Survivor Dec 25 '18 at 13:00
• 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. – Winther Dec 25 '18 at 13:02
• 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? – Maths Survivor Dec 25 '18 at 13:18
• Yes that's right – Winther Dec 25 '18 at 13:19

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).

• "then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence? – Maths Survivor Dec 25 '18 at 13:12
• @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. – Theo Bendit Dec 25 '18 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.

• 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 $\lim n a_n = 0$? – Theo Bendit Dec 25 '18 at 22:44
• Note that $$(2n)a_{2n}\le 2\sum_{n+1}^{2n}a_k\to 0$$since $a_n\ge0$ is monotonic and $\sum_n a_n<\infty.$. – Mark Viola Dec 26 '18 at 4:05