Does the series $\sum (1+n^2)^{-1/4}$ converge or diverge?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

Note that $\sqrt[4]{1+n^2}\le 2n $.

Answer 4

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.