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

alternating-expressionconvergence-divergencereal-analysissequences-and-series

"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.

Best Answer

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:

$$ S_{2N} = \sum_{n=1}^{2N} (-1)^n n a_n = \sum^{2N}_{\substack{n =2\\ n \text{ even}}} \left( -(n-1)a_{n-1} + n a_n\right) \\ = \sum^{2N}_{\substack{n =2\\ n \text{ even}}} a_{n-1} - \sum^{2N}_{\substack{n =2\\ n \text{ even}}} n(a_{n-1}-a_n) =: A_{2N} - B_{2N} \, . $$

On the right we have two sums with non-negative terms, we will show that both sums are bounded above and therefore convergent.

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 $$ a_1, a_2, a_2, a_4, a_4, a_6, a_6 \cdots \, . $$

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