Are $\sum\frac{1}{\sqrt{n(n+1)}},\sum\frac{1}{\sqrt{n(n^2+1)}}$ convergent or divergent? I have tried to answer the Question but I am not sure if the solution for the first sum is alright and I don't know how to handle the other sum. 
The rule of thumb I have is that if we lower the value in the denominator we get a majorant and if we increase the value we get a minorant. I have also tried with root and ratio test, the efforts were fruitless.
$$\frac{1}{\sqrt{n(n+1)}}=\frac{1}{\sqrt{n^2+n}}>\frac{1}{\sqrt{2n^2}}$$
Now I am not sure that if $\zeta(1)=\sum\frac{1}{n}$ diverges (which it does) then also $\sum\frac{1}{\sqrt{2}n}$ must diverge. Am I allowed to say that ? I don't know if I can use distributivity for infinite sums if they do not converge.
About the second sum. I have $\frac{1}{\sqrt{n^3+n}}>\frac{1}{\sqrt{2n^3}}$
However $\zeta(3/2)$ converges. 
Going the other way I have 
$\frac{1}{\sqrt{2n}}>\frac{1}{\sqrt{n^3+n}}$
But $\zeta(1/2)$ diverges
Thank you.
 A: $$\frac{1}{\sqrt{n(n+1)}}\gt\frac{1}{\sqrt{(n+1)^2}}=\frac1{n+1}$$
$$\therefore \sum_{n=1}^\infty\frac{1}{\sqrt{n(n+1)}}\gt\sum_{n=1}^\infty\frac{1}{n+1}=\sum_{n=2}^\infty\frac{1}{n}\to\infty$$
For the other summation
$$\frac{1}{\sqrt{n(n^2+1)}}\lt\frac{1}{\sqrt{n^3}}$$
$$\therefore 0\lt\sum_{n=1}^\infty\frac{1}{\sqrt{n(n^2+1)}}\lt\sum_{n=1}^\infty\frac{1}{\sqrt{n^3}}=\zeta(\frac32)$$
So the summation must converge as it can be bounded and is real.
A: Note that both limits$$\lim_{n\to\infty}\frac{\dfrac1{\sqrt{n(n+1)}}}{\dfrac1n}\text{ and }\lim_{n\to\infty}\frac{\dfrac1{\sqrt{n(n^2+1)}}}{\dfrac1{n^{\frac32}}}$$exist and belong to $(0,\infty)$ (actually, they're both equal to $1$). Therefore your series have the same behaviour as the series$$\sum_{n=1}^\infty\frac1n\text{ and }\sum_{n=1}^\infty\frac1{n^\frac32}$$respectively. That is, the first one diverges, whereas the second one converges.
A: As a complement to the other answers, you are allowed to multiply series by constants and distribute, even if they diverge. If $S_k$ is the sequence of partial sums of the first series, $c$ is the constant multiplier, and $S_k'$ is the sequence of partial sums of the second, then
$$S_k'=cS_k$$
for all $k$. You can always pull constant multiples out of limits, even if they don't exist or are infinite, so if $c\neq 0$ and the first sum diverges, then so does the second. 
