# Problem on inferior limits and real number series [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, kingW3, Shailesh, C. Falcon, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

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.]