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Could you help me prove the two following lemmas on 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 series of positive numbers ( $p_n$ ) increasing 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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closed as too broad by Martin Sleziak, Ben S., Alex Provost, Jonas, Shailesh Apr 1 at 0:17

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

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
Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Mar 31 at 18:27
up vote 1 down vote accepted

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.

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