Solve inequality $\frac{1}{\sqrt {n+1}} + \frac{1}{n+1}>\frac{1}{\sqrt n} - \frac{1}{n}$ for $3\leq n$ I have to show that although the series $\sum_{n=2}^\infty (\frac{(-1)^n}{\sqrt n} + \frac{1}{n})$ is alternating with the absolute value of the terms approaching zero, it does not contradict the test for convergence of alternating series that the series is in fact divergent. 
This must be because the absolute value of the terms are not declining as n approaches infinity. I have trouble showing this though. I see that:
\begin{align*}
|\frac{(-1)^n}{\sqrt n} + \frac{1}{n}| =  \begin{cases} 
      \frac{1}{\sqrt n} + \frac{1}{n} & \text{n is even} \\
      \frac{1)}{\sqrt n} - \frac{1}{n} & \text{n is odd} 
   \end{cases}
\
\end{align*}
so now I could show that if $3\leq n$ and n is odd, then $\frac{1}{\sqrt {n+1}} + \frac{1}{n+1}>\frac{1}{\sqrt n} - \frac{1}{n}$. But I am having trouble showing this. Can anybody help me solve the inequality or suggest another path forward? Thank you very much!
 A: To use the alternating series test, you must prove that $|a_n|>|a_{n+1}|$.  Depending on the sign, there are two inequalities to check:
$$
\frac{1}{\sqrt{n}}+\frac{1}{n}>\left|-\frac{1}{\sqrt{n+1}}+\frac{1}{n+1}\right|
$$
and
$$
\left|-\frac{1}{\sqrt{n}}+\frac{1}{n}\right|>\frac{1}{\sqrt{n+1}}+\frac{1}{n+1}.
$$
These two expressions simplify to checking the inequalities:
$$
\frac{1}{\sqrt{n}}+\frac{1}{n}>\frac{1}{\sqrt{n+1}}-\frac{1}{n+1}
$$
and
$$
\frac{1}{\sqrt{n}}-\frac{1}{n}>\frac{1}{\sqrt{n+1}}+\frac{1}{n+1}.
$$
Moving around the expressions to eliminate negatives, we have to check
$$
\frac{1}{\sqrt{n}}+\frac{1}{n}+\frac{1}{n+1}>\frac{1}{\sqrt{n+1}}
$$
and
$$
\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}}+\frac{1}{n+1}+\frac{1}{n}.
$$
The first inequality is fine because the LHS has $\frac{1}{\sqrt{n}}$ which is greater than $\frac{1}{\sqrt{n+1}}$.  The second inequality fails, however.  To see this, observe that 
$$
\frac{1}{\sqrt{n+1}}+\frac{1}{n+1}=\frac{\sqrt{n+1}+1}{n+1}.
$$
Our goal is to show that this is greater than $\frac{1}{\sqrt{n}}$.  In other words, that 
$$
\frac{\sqrt{n}(\sqrt{n+1}+1)}{n+1}>1.
$$
Since everything is positive, we square the LHS to get
$$
\frac{n(n+2+2\sqrt{n+1})}{(n+1)^2}=\frac{n^2+2n+1-1+2n\sqrt{n+1}}{(n+1)^2}=1+\frac{-1+2n\sqrt{n+1}}{(n+1)^2},
$$
which is greater than $1$ for $n$ sufficiently large.  This contradicts the second inequality.
