# Modifications of the necessary condition for series convergence

Could you help me prove the two following lemmason series convergence?

1) Prove that if $\sum_{n=1} ^{\infty} a_n$ is a series of positive real numbers convergent to $0$, where $(a_n)$ is a monotone convergent sequence, then $(na_n)$ is also convergent.

Why isn't it true for series with arbitrary terms?

If $\sum_{n=1} ^{\infty} a_n$ is convergent ( $a_n >0$ ), can it be that $na_n \rightarrow a \neq 0$ ? (No assumption about monotonicity of $a_n$.)

2) Prove that if $\sum_{n=0} ^{\infty} a_n$ is a convergent seires of positive numbers ( $p_n$ ) incrreasing to infinity, then $\frac{p_0a_0+...+p_na_n}{p_n}$ is convergent to $0$.

As for the first one, I know that from the necessary condition for series convergence we have that for a given $\epsilon <0$ there exists $n_0(\epsilon) = n_0$ such that for $n>n_0$ we have $na_n=ka_n + (n-k)a_n \le \frac{\epsilon}{2} + a_k + ...+ a_n \le \frac{\epsilon}{2} \cdot 2 = \epsilon$

Could you help me with the rest?

Thank you.

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There seems to be something wrong with the first statement. It can't depend on the limit of the series, because that changes when you change a finite number of terms whereas whether $(na_n)$ converges doesn't. Is this a verbatim quote from a book? –  joriki Mar 4 '13 at 10:25
–  Martin Sleziak Dec 26 '13 at 14:44

Let $\displaystyle\sum a_n$ be a slowly convergent alternating series, such as $\displaystyle \sum_1^\infty (-1)^{n+1}\frac{1}{\sqrt{n}}$. The sequence $(na_n)$ is not convergent.
But that's cheating. We produce an example with positive terms. Let $a_n=\dfrac{1}{2^n}$ when $n$ is not a perfect cube. When $n$ is a perfect cube, say $n=k^3$, let $a_n=\dfrac{1}{k^2}$. The series $\displaystyle \sum_1^\infty a_n$ converges, but when $n=k^3$, $na_n$ is large.
If $\displaystyle \sum a_n$ converges, then $(na_n)$ cannot have a non-zero limit $a$. Without loss of generality, we can assume that $a$ is positive. Then for large enough $n$, we have $na_n \gt a/2$, and therefore $a_n \gt \dfrac{a}{2n}$. Because of the divergence of the harmonic series, this implies that $\displaystyle \sum a_n$ diverges.