$\sum a_n$ is convergent $\implies \sum a_n^2$ is convergent?

Question

Assume that $\sum a_n$ is convergent. Does it imply that

I) $n\cdot a_n\rightarrow 0 $ and

Il) $\sum a_n^2$ is convergent?

Are (I) and (II) true if we also assume that the sequence ($a_n$) is monotone?

Answer 1

If $a_n$ is not required to be positive and $\sum a_n$ is not required to converge absolutely then no.

If $|a_n| < |a_{n-1}|$ and $a_{2n} > 0$ while $a_{2n-1} <, 0$ then $\sum_{k=0}^{\infty} a_n$ converge but not necessarily absolutely. $a_n^2$ being positive mean $\sum a_n^2$ must converge absolutely or not at all.

1) $\sum (-1)^n \frac 1n$ converge but $n*(-1)^n\frac 1n = (-1)^n \not \rightarrow 0$.

2) $\sum (-1)^n \frac 1{\sqrt n}$ converge but $\sum [(-1)^n \frac 1{\sqrt n}]^2 = \sum \frac 1n $ diverges.

But if $\sum a_n$ converge absolutely then both are true.

1) $\sum_{k=0}^n |a_k| > \min(|a_0|,......, |a_n|)*n$ so if $n*a_k \not \rightarrow 0$ then this can not converge.

2) If $\sum |a_k| $ converge then all but finite $|a_n| < 1$ so $\sum_{k:|a_k| \ge 1} |a_k| = M$ and $\sum_{k:|a_k| < 1}|a_k| = m$. Then $\sum_{k:|a_k| \ge 1}a_k^2 = N$ and $\sum_{k:|a_k| < 1} a_k^2 < \sum_{k:|a_k| < 1}|a_k| = m$. So $\sum_{k=0}^{\infty} a_k^2 < N + m$. So converges.