Does the series $\sum (1+n^2)^{-1/4}$ converge or diverge? The integral is $\int\left(\,1 + n^{2}\,\right)^{-1/4}\,{\rm d}n$ is not quite possible, so I should make a comparison test.
What is your suggestion?
EDIT:
And what about the series
$$
\sum\left(\, 1 + n^{2}\,\right)^{-1/4} \cos\left(\, n\pi \over 6\,\right)
$$
Does it converge or diverge?
 A: As has been mentioned, $(1+n^2)^{-1/4}\ge\frac1{\sqrt{2n}}$ which diverges by the integral test.
Since $\cos(x+\pi)=-\cos(x)$, we have that
$$
\begin{align}
\sum_{k=n}^{n+11}\cos\left(\frac{k\pi}6\right)
&=\sum_{k=n}^{n+5}\left[\cos\left(\frac{k\pi}6\right)+\cos\left(\frac{k\pi}6+\pi\right)\right]\\
&=\sum_{k=n}^{n+5}\left[\cos\left(\frac{k\pi}6\right)-\cos\left(\frac{k\pi}6\right)\right]\\[9pt]
&=0
\end{align}
$$
That is, the sum of any $12$ consecutive values of $\cos\left(\frac{k\pi}6\right)$ is $0$. Thus, since it has period $12$,
$$
\sum_{k=0}^n\cos\left(\frac{k\pi}6\right)
$$
is bounded independent of $n$. Therefore, because $(1+n^2)^{-1/4}$ monotonically decreases to $0$, the Dirichlet Test says that
$$
\sum_{k=0}^\infty(1+k^2)^{-1/4}\cos\left(\frac{k\pi}6\right)
$$
converges.
A: Note that $\sqrt[4]{1+n^2}\le 2n $.
A: For large $n$ you can approximate $(1+n^2)^{-1/4}\approx \frac{1}{\sqrt{n}}$ which also diverge. Because of $\sum \frac{1}{n}\le \sum \frac{1}{\sqrt{n}}$ does.
Regarding $$\sum (1+n^2)^{-1/4} \cos {\frac{n\pi}{6}}$$
With $(1+n^2)^{-1/4}\to 0$ and $\cos {\frac{n\pi}{6}}$ is oscillating around zero. Hint: Look what you get for $\sum_{n=1+m}^{12+m} \frac{1}{\sqrt{n}}\cos\frac{n\pi}{6}$ approximately for all $m$ s.
A: For the second part, notice that there is a pattern to the factor of $\cos\left(\frac{\pi n}{6}\right)$ which repeats every 12 terms. Use this pattern to think of the sequence as alternating.
