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I've been trying to solve these two problems regarding limits and series, yet I couldn't find a proper solution:

Let $\sum_{n=0}^\infty x_n$ be a real number series, which is convergent. Show that $\mathop{\underline{\lim}}_{n \to \infty} nx_n = 0$. Additionally, knowing that $x_{n+1} < x_n$, find that $\exists \ lim_{n\rightarrow \infty}nx_n = 0$.

Any help with these two will be much appreciated. Thank you all in advance!

If someone could give step-by-step proof, that would be great.

PS: I've asked this problem in another one of my questions, but Mr. Henning Makholm https://math.stackexchange.com/users/14366/henning-makholm has suggeted that I should ask them separately.

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closed as off-topic by Did, kingW3, Shailesh, C. Falcon, Daniel W. Farlow Feb 4 '17 at 3:20

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a) The first statement is not true. $\sum_n \frac{(-1)^n}{n}$ is convergent (by the Leibnitz-test) but $\liminf_n (-1)^n = -1 \neq 0$.

b) If additionally $x_{n+1}<x_n$ for all $n\in \mathbf N$ the second statement is true. If $\sum_n x_n$ is convergent then $s_n := \sum_{j=1}^n x_j$ is a Cauchy sequence. Hence $$s_{2n}-s_n= \sum_{j=n+1}^{2n} x_j < \varepsilon \qquad (\star)$$ where $\varepsilon >0$ is arbitrary. Therefore we get (since $(x_n)_n$ is decreasing) $\sum_{j=n+1}^{2n} x_j > nx_{2n}$ what shows $\lim_n nx_n =0$ (since $\varepsilon $ was arbitrary). [For the last conclusion you may have a look again at $(\star)$ since we do not have to pick $m=2n$ and $n$. We can choose for example $m=0$ and $m=1$ to get $2nx_{2n} < \varepsilon$ and $(2n+1)x_{2n+1}< \varepsilon$. The rest should be easy for you.]

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