Convergence of the series $ \sum_{n=1}^{\infty} \arctan(\frac{1}{1 + n + n^{\alpha}}) $

Question

For which $\alpha > 0 $ does the series

$$ \sum_{n=1}^{\infty} \arctan(\frac{1}{1 + n + n^{\alpha}}) $$

converge? I suspect it does for $\alpha > 1$ but how do I show that and that it doesn't converge for $\alpha \leq 1$?

Answer 1

Use L'Hospital's Rule to show that $\frac {\arctan x -x} {x^{2}} \to 0$. Add and subtract $\frac 1 {1+n+n^{\alpha }}$ in the general term of the series and observe that $\sum \frac 1 {(1+n+n^{\alpha })^{2}} < \infty$ for all $\alpha >0$ (because $1+n+n^{\alpha } >n$). Hence the original series converges iff $ \sum \frac 1 {1+n+n^{\alpha }}$ converges which is true iff $\alpha >1$.

Answer 2

You have $\arctan(x) \sim x$ as $x \to 0$ so $$\arctan(1/(1 + n + n^\alpha)) \sim {1\over 1 + n + n^\alpha}$$ as $n\to\infty$. Break out the limit comparison test.

Answer 3

From the Taylor series about $x=0$ ($\arctan(x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$) or other means, for $0\leq x \leq 1$

$$x - x^3/3 \leq \arctan(x) \leq x$$

Hence for any $n> 0$,

$$\frac{1}{1+n+n^{\alpha}} - \frac{1}{3(1+n+n^{\alpha})^3} \leq \arctan\left( \frac{1}{1+n+n^{\alpha}}\right) \leq \frac{1}{1+n+n^{\alpha}}$$

Now, clearly $$n^{\max\{1,\alpha\}} <1+n^{\max\{1,\alpha\}} < 1 + n+n^\alpha \leq 1 + 2n^{\max\{1,\alpha\}} \leq 3n^{\max\{1,\alpha\}}$$

Applying this to the above bounds on $\arctan \frac{1}{1+n+n^{\alpha}}$, we get

$$ \frac{1}{3n^{\max\{1,\alpha\}}} - \frac{1}{3 n^{3\max\{1,\alpha\}}} < \arctan\left( \frac{1}{1+n+n^{\alpha}}\right) \leq\; \frac{1}{3n^{\max\{1,\alpha\}}}$$

If $\alpha \leq 1$ then $$\sum_{n=1}^{\infty}\left(\frac{1}{3n^{\max\{1,\alpha\}}} - \frac{1}{3 n^{\max\{3,3\alpha\}}}\right) = \frac13\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}}_{\to \infty} - \frac13\underbrace{\sum_{n=1}^{\infty} \frac{1}{n^3}}_{<\infty}$$ hence by the comparison test, $\sum_{n=1}^{\infty}\arctan\left( \frac{1}{1+n+n^{\alpha}}\right)$ diverges.

If $\alpha > 1$ then $$\sum_{n=1}^\infty \frac{1}{3n^{\max\{1,\alpha\}}} =\frac13\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}}_{<\infty}$$ hence $\sum_{n=1}^{N}\arctan\left( \frac{1}{1+n+n^{\alpha}}\right)$ is bounded above and given that all terms are non-negative $\sum_{n=1}^{\infty}\arctan\left( \frac{1}{1+n+n^{\alpha}}\right)$ exists.