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Let $\sum_{n=1}^{\infty}{a_n}$ be a sequence such that $a_n > 0$ for all $n$ and

$\sum_{n=1}^{\infty} a_{n}$ converges. Decide if $\sum_{n=1}^{\infty} a^2_n$ converges of diverges.

I simply put $\sum_{n=1}^{\infty} a^2 = \frac{1}{6}(n+1)(2n+1)$ and when n approaches infinity, so does the sum, so my conclusion is that the series diverges.

I'm not sure if this is the right way to approach the problem though, as I'm not using the information given at the beginning: Where does the convergence of $\sum_{n=1}^{\infty} a_n$ come into play? By comparison $a_n < a^2_n$ , but that doesn't lead me anywhere, as it would have to be opposite for the test to work, right?

novo
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    No, it is absolutely not the correct approach. It seems that you've confused the finite summation formula $\sum_{i = 1}^n i^2 = n(n + 1)(2n + 1)/6$ with something about geometric series. You also have the inequality $a_n < a_n^2$, which is not typically true. Also, your title doesn't have anything to do with your question. –  Oct 20 '17 at 15:59

2 Answers2

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$\sum_\limits{n=1}^\infty a_n$ converges.

$\lim_\limits {n\to\infty} a_n = 0$

$\exists N>0:n>N $$\implies |a_n| <1\\ \implies |a_n^2| < |a_n|$

$\sum_\limits{n=1}^\infty a_n^2$ converges by the comparison test.

Doug M
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Since $\sum a_n$ converges, $lim_na_n=0$, there exists $N$ such that $n>N$ implies $0<a_n<1$, this implies that $a_n^2<a_n$. Use the comparison theorem to show that $\sum a_n^2$ converges.

Tsemo Aristide
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