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


(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.

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

1 Answer 1

  • For a) I would have written :

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.

  • For b), well done
  • For c), it's true

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.

  • For d), it's fine
  • For e), good too

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