Test series for convergence: $\sum\limits_{n=0}^{\infty} (-1)^n \frac{\sqrt{n}}{n+1}$ Test the series for convergence $\sum\limits_{n=0}^{\infty} (-1)^n \frac{\sqrt{n}}{n+1}$.
By using the Leibniz criterion, I'd have to show that the sequence $a_n = \frac{\sqrt{n}}{n+1}$ is monotonous and that it converges to $0$.
For the limit, I have $\lim\limits_{n \to \infty} \frac{\sqrt{n}}{n+1} = \lim \frac{\frac{\sqrt{n}}{n}}{\frac n n + \frac 1 n} = \frac 0 1 = 0$.
Is there any other step I could put in-between to make it more obvious or correct?
However, for the monotony, I'm kind of stuck. I've tried by induction, but I can't see how to get from $a_{n+1} \le a_n$ to $a_{n+2} \le a_{n+1}$
 A: Your series converges by Leibniz test.
Indeed, Setting the function $$f(x) = \frac{\sqrt{x}}{x+1} \implies f'(x) =\frac{1-x}{2\sqrt{x}(x+1)^2} \le 0~~for ~~x\ge 1$$
Then, the sequence $${n \mapsto} \frac{\sqrt{n}}{n+1} \qquad \text{is stictly decreasing } $$ and and we also have, 
 $$\lim\limits_{n \to \infty} \frac{\sqrt{n}}{n+1}  = 0\qquad \text{ } $$
Hence your series $$\sum\limits_{n=0}^{\infty} (-1)^n \frac{\sqrt{n}}{n+1}$$converges by Leibniz test.
A: In order to prove that $n\mapsto \frac{\sqrt{n}}{n+1}$ is decreasing on $\mathbb{N}^+$, it is enough to prove that $g:n\mapsto\sqrt{n}+\frac{1}{\sqrt{n}}$ is increasing. Since $g(n) = 2\cosh\left(\frac{1}{2}\log n\right)$ that is trivial by composition of non-negative and increasing functions on $\mathbb{R}^+$. It follows that $\sum_{n\geq 1}\frac{\sqrt{n}}{n+1}(-1)^n$ is (conditionally) converging by Leibniz' test.
A: There's no induction needed here -- i.e., you don't need to get from $a_{n+1}\le a_n$ to $a_{n+2}\le a_{n+1}$.  Instead, you just need to prove that $a_{n+1}\le a_n$, which is straightforward: Since all the $a_n$'s are non-negative, we have the string of equivalences
$$\begin{align}
{\sqrt{n+1}\over(n+1)+1}\le{\sqrt n\over n+1}
&\iff{n+1\over(n+2)^2}\le{n\over(n+1)^2}\\
&\iff(n+1)^2\le n(n+2)^2\\
&\iff n^3+3n^2+3n+1\le n^3+4n^2+4n\\
&\iff 1\le n^2+n
\end{align}$$
The final inequality is clearly true for $n\ge1$, hence so is the inequality $a_{n+1}\le a_n$ (for $n\ge1$).  
Remark:  It would make sense to start the summation with $n=1$ rather than $n=0$, since $a_0=0$.
A: 1)Limit:
$a_n := √n/(n+1).$
$0 \lt √n/(n+1) \lt √n/n = 1/√n.$
$\lim_{n \rightarrow \infty} a_n \le \lim_{n\rightarrow \infty }(1/√n) =0.$
2) Monotonicity:
$a_n = √n/(n+1) = \dfrac{1}{√n + 1/√n)}$
is strictly decreasing.
Consider: $ f(x)= x +1/x, x >1.$
$f'(x) = 1 -1/(x^2) >0$, I.e 
$f(x)$ strictly increasing.
$\rightarrow:$
$[f(x)]^{-1}$ strictly decreasing.
Can you take it from here?
A: For the limit, equivalents are much shorter: $n+1\sim_\infty n$, so
$$\frac{\sqrt n}{n+1}\sim_\infty\frac{\sqrt n}{n}=\frac1{\sqrt n}\to 0.$$
\begin{align}
&\text{For the monotony: }& a_{n+1}<a_n&\iff a_{n+1}^2=\frac{n+1}{(n+2)^2}<a_n^2=\frac  n{(n+1)^2}&\qquad\qquad&\\
&&&\iff (n+1)^3<n(n+2)^2\iff1<n^2+n.
\end{align}
A: Let $f(x)=\frac{\sqrt x}{x+1}$. Then $a_n=f(n)$. Since
$$ f'(x)=\frac{1-x}{2\sqrt x(x+1)}\le 0 $$
$\{a_n\}$ is decreasing. Noting $ \lim_{x\to\infty}f(x)=0$, one obtains that $\sum_{n=1}^\infty(-1)^na_n$ converges.
