# If $(a_n)$ is a decreasing real sequence and $\sum a_n$ converges, then does $\sum (-1)^n n a_n\$ converge?

"Motivation"/Introduction:

1. If $$(a_n)$$ is a decreasing real sequence and $$\displaystyle\sum a_n$$ converges, then $$n a_n \to 0,\$$ for example, by the Cauchy Condensation test.

2. If $$(a_n)$$ is a real sequence and $$\displaystyle\sum a_n$$ converges but $$(a_n)$$ is not necessarily decreasing, then does $$n a_n\to 0?$$ No: for example, take:

$$a_n:= \begin{cases} 2^{-n}&\text{if}\, n\neq 2^k\\ \frac{1}{k^2}&\text{if}\, n = 2^k\\ \end{cases}$$

Therefore, if $$(a_n)$$ is a real sequence and $$\displaystyle\sum a_n$$ converges but $$(a_n)$$ is not necessarily decreasing, then $$\displaystyle\sum n a_n\$$ and $$\displaystyle\sum (-1)^n n a_n\$$ do not necessarily converge, by the counter-example above.



This leads to the following question, for which I do not have an answer:

If $$(a_n)$$ is a decreasing real sequence and $$\displaystyle\sum a_n$$ converges, then does $$\displaystyle\sum (-1)^n n a_n\$$ converge?

I was thinking we could apply Dirichlet's test or Cauchy's Condensation test somehow, but I don't quite see how.

Yes, $$\sum (-1)^n n a_n$$ is convergent under the given conditions.
Since $$na_n \to 0$$ it suffices to show that the partial sums with even index are convergent:
$$S_{2N} = \sum_{n=1}^{2N} (-1)^n n a_n = \sum^{2N}_{\substack{n =2\\ n \text{ even}}} \left( -(n-1)a_{n-1} + n a_n\right) \\ = \sum^{2N}_{\substack{n =2\\ n \text{ even}}} a_{n-1} - \sum^{2N}_{\substack{n =2\\ n \text{ even}}} n(a_{n-1}-a_n) =: A_{2N} - B_{2N} \, .$$
The first one is easy: $$A_{2N} = \sum^{2N}_{\substack{n =2\\ n \text{ even}}} a_{n-1} \le \sum_{n=1}^\infty a_n \, .$$ The second sum can be estimated by adding the terms for odd $$n$$: $$B_{2N} \le \sum_{n=2}^{2N} n(a_{n-1}-a_n) = \sum_{n=2}^{2N} \bigl((n-1)a_{n-1} - na_n \bigr) + \sum_{n=2}^{2N} a_{n-1}$$ There is a telescoping sum on the right, so we get $$B_{2N} \le a_1 - 2Na_{2N} + \sum_{n=2}^{2N} a_{n-1} \le a_1 + \sum_{n=1}^\infty a_n$$ and that finishes the proof.
Remark: Using the above estimates we can also obtain lower and upper bounds for the value of $$\sum_{n=1}^\infty (-1)^n n a_n$$ in terms of the given series $$\sum_{n=1}^\infty a_n$$. Let $$S_{\text{odd}} = \sum^{\infty}_{\substack{n =1\\ n \text{ odd}}} a_n \, , \, S_{\text{even}} = \sum^{\infty}_{\substack{n =2\\ n \text{ even}}} a_n \, .$$ Then $$A_{2N} \to S_{\text{odd}}$$ and $$0 \le B_{2N} \le a_1 +S_{\text{odd}} + S_{\text{even}}$$, so that $$\boxed{-a_1 - S_{\text{even}} \le \sum_{n=1}^\infty (-1)^n n a_n \le S_{\text{odd}} \, .}$$ Both bounds are sharp, as can be seen by considering sequences of the form $$a_1, a_1, a_3, a_3, a_5, a_5, \cdots$$ and $$a_1, a_2, a_2, a_4, a_4, a_6, a_6 \cdots \, .$$