Calc 2 Limit Comparison Test I need to find out if it converges.
$$\sum\limits_{k=0}^{+\infty} \frac{\sqrt{k}}{(k^3+1)}$$
The solutions manual says to use $k^{-5/2}$ as a model case to use for the limit comparison test.
I'm not sure how one would be able to come up with that.
 A: $$\sum\limits_{k=0}^{\infty} \frac{\sqrt{k}}{k^3+1}=\sum\limits_{k=1}^{\infty} \frac{\sqrt{k}}{k^3+1}$$
First note that
$$\left|\frac{\sqrt{k}}{k^3+1}\right|\leq\left|\frac{\sqrt{k}}{k^3}\right|=\left|\frac{1}{k^{\frac52}}\right|$$
Also since $\frac52\gt 1$, we have
$$\zeta\left(\frac52\right)=\sum\limits_{k=1}^{\infty} \left|\frac{1}{k^{\frac52}}\right|=\mbox{convergent}$$
Now lets use the limit comparison test
$$\lim\limits_{k\to\infty}\frac{\left|\frac{\sqrt{k}}{k^3+1}\right|}{\left|\frac{1}{k^{\frac52}}\right|}=\lim\limits_{k\to\infty}\left|\frac{k^3}{k^3+1}\right|$$
$$=\lim\limits_{k\to\infty}\left|\frac{1}{1+\frac{1}{k^3}}\right|=\left|\frac{1}{1+0}\right|=1$$
Therefore by the limit comparison test,
$$\sum\limits_{k=1}^{\infty} \left|\frac{\sqrt{k}}{k^3+1}\right|=\mbox{convergent}$$
Which implies that
$$\sum\limits_{k=1}^{\infty} \frac{\sqrt{k}}{k^3+1}=\mbox{absolutely convergent}$$
A: As $k \to \infty$, top looks like $k^{1/2}$ and bottom looks like $k^3$, so their ratio is $1/k^{5/2}$.
