Infinite Series Convergence Verification I'm currently tackling the following infinite series in order to determine if it converges or not:
$$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n}$$
What I've done is rearranged this to make use of the comparison test, shown below:
$$= \frac{\sqrt{n+1}-\sqrt{n}}{n} * (\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})$$
$$= \frac{(n+1)-(n)}{n(\sqrt{n+1}+\sqrt{n})}$$
$$= \frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$
$$= \frac{1}{n^{3/2}+\sqrt{n+1}} < \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} $$
However, $\frac{1}{n}$ is smaller than $\frac{1}{\sqrt{n}}$, therefore:
$$  \frac{1}{n}   < \frac{1}{\sqrt{n}}$$ 
This implies divergence by the comparison test due to the fact $\frac{1}{n}$ diverges, and if a smaller series diverges any larger series must as well.
Is this right? If I've made any mistakes in any part, please tell me!
 A: You are correct that the $n$th term of your sequence is less than $\frac{1}{\sqrt{n}}$, but you cannot conclude that the series diverges. After all, $\frac{1}{n^2}<\frac{1}{\sqrt{n}}$ and $\sum\frac{1}{n^2}$ converges. To use the comparison test to show that a series diverges, you need a comparison in the opposite direction.
Instead, note that
$$ \frac{1}{n^{\frac{3}{2}}+n\sqrt{n+1}}<n^{-\frac{3}{2}}$$
and $\sum n^{-\frac{3}{2}}$ converges.
A: $$u_n=\frac {1}{\sqrt {n}}(\sqrt {1+\frac {1}{n}}-1) $$
$$(1+X)^a=1+aX (1+\epsilon (X)) $$
with $\lim_0\epsilon (X)=0$
or
$$(1+X)^a-1\sim aX \;\;(X\to 0) $$
thus
$$\sqrt {1+1/n}-1\sim \frac {1}{2n}\;\;(n\to+\infty) $$
finally
$$u_n\sim \frac {1}{2n^\frac 32} \;\;(n\to+\infty) $$
and
$\sum u_n $  converges.
A: In Italy we would say a'voja.
Let $a_n=\sqrt{n+1}-\sqrt{n}$ and $b_n=\frac{1}{n}$. By summation by parts,
$$ \sum_{n=1}^{N}a_n b_n = \frac{\sqrt{N+1}-1}{N}+\sum_{n=1}^{N-1}\frac{\sqrt{n+1}-1}{n(n+1)}\tag{1} $$
hence:
$$ \lim_{N\to +\infty}\sum_{n=1}^{N}a_n b_n = -1+\color{blue}{\sum_{n\geq 1}\frac{1}{n\sqrt{n+1}}}\tag{2} $$
where the blue series is absolutely convergent by the $p$-test. For any $n\geq 1$ we have that $\frac{1}{n\sqrt{n+1}}$ is extremely close to $\frac{2}{\sqrt{n-\frac{1}{6}}}-\frac{2}{\sqrt{n+\frac{5}{6}}}$, hence the value of the original series is close to $-1+\frac{2}{\sqrt{\frac{5}{6}}}$ by creative telescoping.
$$ \sum_{n\geq 1}\frac{\sqrt{n+1}-\sqrt{n}}{n}\approx\color{blue}{\frac{6}{5}}.\tag{3}$$
A: Hint: $\dfrac{1}{n (\sqrt {n}+\sqrt {n+1})}\le\dfrac {1}{n^{3/2}}$ and $\frac{3}{2}>1$. 
