$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
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.
Best Answer
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.