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Considering $ x \in \mathbb{R} \setminus\{0\}$, I want to study: $$ \sum_{n=1}^{\infty} \frac{(-1)^n}{x^n}n\left(\frac{\pi}{2}-\arctan(n)\right)\log\left(2+\frac{1}{n}\right)$$

I want to find for which the set of $x$ where the series converge and whether to not it converges uniformly on $(1,\infty)$.

I was trying to use Mertens theorem for product of series, without success because the series with only $\frac{\pi}{2}-\arctan(n)$ diverges. What should I try?

egreg
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qcc101
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  • The ratio test yields the limit $1/|x|$, so you have absolute convergence for $|x|>1$ and no convergence for $0<|x|<1$. – egreg Feb 28 '18 at 22:57

1 Answers1

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Note that $$\lim_{x\to\infty}\frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}=\lim_{x\to\infty}\frac{-\frac{1}{1+x^2}}{-\frac{1}{x^2}}=1,$$ therefore $n\left(\frac{\pi}{2}-\arctan n\right)\to 1$ as $n\to\infty$. Moreover, $\log\left(2+\frac{1}{n}\right)\to\log 2$, which means that for the general term of the series to converge to $0$, we should have that $(-1)^nx^{-n}\to 0$, therefore $|x|>1$. On the other hand, the series converges absolutely when $|x|>1$, so the series converges if and only if $|x|>1$.

For uniform convergence, note that the series cannot be uniformly Cauchy (check the behavior close to $1$), so the series is not uniformly convergent in $(1,\infty)$.

detnvvp
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  • I understood your reasoning up until the uniform convergence. Can you explain it a bit further? – qcc101 Mar 01 '18 at 08:17
  • It turns out that this is not straightforward. I'll take a look to see how to handle it. – detnvvp Mar 01 '18 at 16:30
  • I did it in another way, calculating $\lim_{n\to \infty}sup(f_n(x) - f(x))$. With f(x) = $\lim_{n\to \infty}f_n(x)$ = 0, the sup is 1 which is not 0 and so there is no uniform convergence. Is that right? – qcc101 Mar 01 '18 at 17:05