# Assertions about convergence/divergence of infinite series

$$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$ I am self-learning Real Analysis from Understanding Analysis by Stephen Abbot. I'd like someone to verify if my proofs/counterexamples about the below assertions are fine.

Consider each of the following propositions. Provide short proofs for those that are true and counterexamples for any that are not.

(a) If $$\sum a_n$$ converges absolutely, then $$\sum a_n^2$$ also converges absolutely.

(b) If $$\sum a_n$$ converges and $$(b_n)$$ converges, then $$\sum a_n b_n$$ converges.

(c) If $$\sum a_n$$ converges conditionally, then $$\sum n^2 a_n$$ diverges.

(d) If $$\sum a_n$$ with $$a_n > 0$$ is convergent, then is $$\sum \sqrt{a_n}$$ always convergent? Either prove it or give a counterexample.

(e) If $$\sum a_n$$ with $$a_n > 0$$ is convergent, then is $$\sum \sqrt{a_n a_{n+1}}$$ always convergent. Either prove it or give a counterexample.

Proof.

(a) We know that, for real numbers $$a,b \in \mathbf{R}$$ \begin{align*} \absval{a}^2 + \absval{b}^2 \le (\absval{a} + \absval{b})^2 \end{align*}

So, we can write, \begin{align*} \absval{a_{m+1}}^2 + \ldots + \absval{a_{n}}^2 \le (\absval{a_{m+1}} + \ldots + \absval{a_{n}})^2 \end{align*}

As $$\sum a_n$$ is absolutely convergent, by the Cauchy criterion, given any $$\epsilon > 0$$, there exists $$N \in \mathbf{N}$$, such that \begin{align*} \absval{\absval{a_{m+1}} + \ldots + \absval{a_{n}}} < \sqrt{\epsilon} \end{align*}

for all $$n > m \ge N$$. Therefore,

\begin{align*} \absval{a_{m+1}}^2 + \ldots + \absval{a_{n}}^2 &\le (\absval{a_{m+1}} + \ldots + \absval{a_{n}})^2\\ &<(\sqrt{\epsilon})^2 = \epsilon \end{align*}

Thus, $$\sum a_n^2$$ is absolutely convergent.

(b) This proposition is false. As a counterexample, consider $$a_n := \frac{(-1)^{n+1}}{\sqrt{n}}$$ and $$b_n := \frac{(-1)^{n+1}}{\sqrt{n}}$$. $$\sum a_n$$ and $$(b_n)$$ are convergent, but $$\sum a_n b_n$$ is the harmonic series, which is well-known to diverge.

(c) I was not able to find any counter-examples. For example, if we define $$a_n := \frac{(-1)^n}{n}$$ then the series $$\sum n^2 a_n$$ diverges.

(d) No. As a counterexample, consider $$a_n = \frac{1}{n^2}$$. $$\sum a_n$$ is convergent, but $$\sum \sqrt{a_n}$$ is divergent.

(e) The geometric mean of two numbers is always less than or equal to the arithmetic mean. \begin{align*} \sqrt{a_n a_{n+1}} \le \frac{a_n + a_{n+1}}{2} \end{align*}

Both $$(a_n)$$ and $$(a_{n+1})$$ are convergent, and hence $$\sum (a_n + a_{n+1})/2$$ is also convergent by the algebraic limit theorem.

By the Comparison test, we must have $$\sum \sqrt{a_n a_{n+1}}$$ convergent.

• what does converge conditionally means?
– math
Jan 1, 2021 at 17:41
• $\sum a_n$ converges conditionally, if and only if, the series is convergent, but it is not absolutely convergent. Jan 1, 2021 at 17:42
• For c), if $\sum n^2a_n$ converges, then $n^2a_n\rightarrow0$; so eventually $|a_n|<1/n^2$. Jan 1, 2021 at 17:43

Since $$\sum |a_n|$$ converges, there exists a rank $$N$$ such that $$\forall n\geqslant N$$, $$|a_n|<1$$. Then $$\forall n\geqslant N, |a_n|^2 \leqslant |a_n|$$. By theorem of comparison between positive terms series, we get that $$\sum |a_n|^2$$ converges.
You can prove it by contraposition. If $$\sum n^2 a_n$$ converges then $$n^2 a_n \rightarrow 0$$ so there exists a rank $$N$$ such that $$\forall n\geqslant N, n^2 |a_n| < 1$$, so $$\forall n\geqslant N, |a_n|<\frac{1}{n^2}$$. By comparison, the serie converges absolutely.