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I don't know how should I determine if this series is convergent or divergent. $$\sum_{n=1}^\infty\frac{e^{\arctan(n)}}{n^2+1}$$

Any help would be appreciated.

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  • $\begingroup$ How big or small can $\arctan n$ be? $\endgroup$ – user121926 Jan 17 '14 at 16:23
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    $\begingroup$ Hint: $\arctan n$ tends to $\pi/2$ as $n$ tends to $\infty$. $\endgroup$ – David Mitra Jan 17 '14 at 16:23
  • $\begingroup$ yeah, exactly, as n -> (Infinite) , arctang - > (PI/2), but I don't understand how this can help. $\endgroup$ – bossModus Jan 17 '14 at 16:25
  • $\begingroup$ What is the biggest or smallest that $e^{\arctan n}$ can be? $\endgroup$ – user121926 Jan 17 '14 at 16:25
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It is convergent as arctan(x) for very large x, converges to $\frac{\pi}{2}$. Thus the series is bounded by $$e^{\frac{\pi}{2}}\sum\limits_{k=1}^{\infty}\frac{1}{k^2+1}$$ which is convergent. To prove this consider $\sum\limits_{k=1}^{\infty}\frac{1}{k^2}$ and to prove that this converges, consider the integral $\int\limits_{1}^{\infty}\frac{1}{x^2}$ in relation to the sum.

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Since $\arctan(x) < \frac{\pi}{2}$ for $x \ge 1$, we have that $e^{\arctan(n)} < e^{\pi/2}$ for $n \ge 1$. Thus: $$\sum_{n=1}^\infty\frac{e^{\arctan(n)}}{n^2+1}\le \sum_{n=1}^\infty\frac{e^{\pi/2}}{n^2+1} = e^{\pi/2}\sum_{n=1}^\infty\frac{1}{n^2+1}$$

Therefore, by comparison, the series...

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How about using the integral test?

$$\int_1^\infty \frac{e^{\arctan x}}{x^2+1}$$ $$u=\arctan x,\;du=\frac{1}{x^2+1}$$ $$\int e^udu=e^u+C=e^{\arctan x}+C$$ $$\lim_{b\rightarrow\infty} e^{\arctan x}|_1^b=\lim_{b\rightarrow\infty}(e^{\arctan b}-e^{\arctan 1})=e^{\frac{\pi}{2}}-e^{\frac{\pi}{4}}$$

Since the integral converges, the series converges.

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  • $\begingroup$ I like this answer, but I have a question. How can we prove that if the integral is bounded, the summation would be convergence as well? $\endgroup$ – Rasa Jan 17 '14 at 17:08
  • $\begingroup$ Not going to copy it, but there's a proof here: en.wikipedia.org/wiki/Integral_test_for_convergence#Proof $\endgroup$ – Nick D. Jan 17 '14 at 17:59
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First note that this sum produces a monotonic sequence since all of its terms are non-negative so all that remains to show is that the sequence is bounded and that will imply convergence (Monotone Convergence Theorem). We can show it's bounded by comparing it to another series that is bounded.

Since $\frac{-\pi}{2} \leq \arctan(n) \leq \frac{\pi}{2}$ you can see by comparison that since $\sum_{n=1}^{+\infty}{\frac{e^{\pi/2}}{n^2+1}}$ converges and $\arctan(n) \leq \frac{1}{n^2+1}$ for $1 \leq n \leq +\infty$ then it also converges.

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