Multiple choice question on sequence and series

Question

$\{xn\}$ be a sequence of positive real numbers such that $\sum_{n=1}^{\infty} x_n$ converges. which of following are true

1. The series $\sum_{n=1}^{\infty} \sqrt{x_nx_{n+1}}$ converges
2. $lim_{n\rightarrow \infty}nx_n =0$
3. The series $\sum_{n=1}^{\infty} sin^2 x_n $ converges
4. $\sum_{n=1}^{\infty} \frac{\sqrt{x_n}}{1+\sqrt{x_n}}$ converges

I think first can be proved by limit comparison test. For two by divergence test, $lim_{n\rightarrow \infty}x_n =0$, but counter example for the given statement I dont have. 4 by direct comparison convergent.

Answer 1

1. True. By the AM-GM, $\sqrt{x_n x_{n+1}} \leq \frac{x_n + x_{n+1}}{2}$, so \begin{align*} \sum_{n=1}^\infty \sqrt{x_n x_{n+1}} &\leq \sum_{n=1}^\infty \frac{x_n + x_{n+1}}{2} \\ &= \frac{1}{2} \sum_{n=1}^\infty x_n + \frac{1}{2} \sum_{n=2}^\infty x_n \text{,} \end{align*} so converges.
2. False. Answered elsewhere.
3. True. Since $\sum x_n$ converges, $x_n \rightarrow 0$. So there is an $N > 0$ such that for all $n > N$, $x_n < 1$. Then, for $n > N$,
   $$ 0 \leq \sin^2 x_n \leq x_n^2 < x_n $$ (... because $\sin'(x) \leq 1$ and $\sin''(x) < 0$ for $x \in [0,1]$). By the comparison test, since $\sum_{n=1}^\infty x_n$ converges, so does $\sum_{n=1}^\infty \sin^2 x_n$.
4. False. Consider $x_n = n^{-3/2}$. Then $$ \frac{\sqrt{x_n}}{1+\sqrt{x_n}} = \frac{n^{-3/4}}{1+n^{-3/4}} \cdot \frac{n^{3/4}}{n^{3/4}} = \frac{1}{n^{3/4}+1} \text{,} $$ which is greater than $n^{-1}$ for sufficiently large $n$. (In fact, for $n \geq 4$.)

Answer 2

1) $\displaystyle\sum\sqrt{x_{n}x_{n+1}}\leq\left(\sum x_{n}\right)^{1/2}\left(\sum x_{n+1}\right)^{1/2}$.

2) Consider $a_{n}=1/k^{2}$ for $n=k^{4}$ and $a_{n}=1/n^{2}$ otherwise, $(na_{n})$ has no limit.

3) $\sin^{2}x_{n}=\dfrac{\sin^{2}x_{n}}{x_{n}^{2}}x_{n}^{2}\leq x_{n}^{2}\leq x_{n}$ for $n$ large.