# Calculating $\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$

I am studying for my exam in real analysis and I am having difficulties with some of the material, I know that the following should be solved by using the counting measure and LDCT, but I don't know how.

For $\alpha>0,$ Calculate $$\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$$

I would greatly appreciate it if in the answers you can include all the details about the theorems used (since there is a high importance for the arguments for why we can do what we do in each step, and I don't understand the material good enough to be able to understand that some step is actually not trivial and uses some theorem)

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By "arctg", do you by any chance mean "arctangent", which can be written either as "arctan", "atan", or "$\tan^{-1}$"? (someone has edited it to say that. Is it what you intended?) – Glen O Apr 1 '13 at 12:45
@GlenO - Yes, this is what I mean. thanks for asking it to clarify, and sorry for the wrong spelling of the name of the function – Belgi Apr 1 '13 at 12:48

Note that for terms of serie $\sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$ we have \begin{align} \left|\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right| = & \left| \frac{\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right| \\ \leq & \left| \frac{2\pi}{n^{\alpha}+k^{3/2}}\right| \\ = & \left| \frac{\pi}{\frac{n^{\alpha}+k^{3/2}}{2}}\right| \\ \leq & \frac{\pi}{\sqrt[2\,]{n^{\alpha}\cdot k^{3/2}}} \\ = & \frac{\pi}{n^{\frac{\alpha}{2}}\cdot k^{3/4}} \\ = & \frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}} \end{align} Then $$0\leq \left| \sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right|\leq \sum_{k=1}^n\left|\frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}}\right|$$ By Squeeze Theorem for Sequences, $\sum_{k=1}^n\left|\frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}}\right|\to 0$ (for all $\alpha>0$) implies $\left|\sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right| \to 0$

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Can you please explain the second inquality and the last line: what is the parameter that tends to infinity in the beginning and the "implies" part – Belgi Apr 1 '13 at 13:32
@Belgi $|\arctan(x)|\leq \frac{\pi}{2}\leq 2\pi$. Remember of graphic of function $\arctan(x)$. – MathOverview Apr 1 '13 at 13:33
@Belgi, Sorry. The second inequality is aritimetic-geometric inequality: $\sqrt[2\,]{a\cdot b}\leq \frac{a+b}{2}$ implies $\frac{1}{\frac{a+b}{2}}\leq \frac{1}{\sqrt[2\,]{a\cdot b}}$ – MathOverview Apr 1 '13 at 13:36
Perhaps I'm deeply wrong, but $\sum^{n}_{k=1}\biggr|\frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}}\biggr|=(‌​\pi/n^{\alpha/2})\sum^{n}_{k=1}\biggr|\frac{1}{k^{3/4}}\biggr|$. Assuming, say, $\alpha=0.00001$, we have $\frac{\pi}{n^{\alpha/2}}\sum^{n}_{k=1}\biggr|\frac{1}{k^{3/4}}\biggr|\to\infty$ as $n\to\infty$. – Johnny Westerling Apr 1 '13 at 16:30
Indeed, the series with respect ot $k$ has to be evaluated in full, independently on $n$. So, calling $N$ the upper limit, we first have to send $N\to\infty$ and then we can send $n\to\infty$. Unfortunately, when sending $N\to\infty$ we get a divergent series, so we cannot conclude anything. – bartgol Mar 14 at 20:36

Let $a(n,k):=(-1)^k\frac{\arctan(n^2k)}{n^{\alpha}+k^{3/2}}$; then $|a(n,k)|\leqslant \frac{\pi}{2k^{3/2}}$ for each $n$, hence $$\left|\sum_{k=1}^{+\infty}a(n,k)\right|\leqslant \sum_{k=1}^N|a(n,k)|+\frac{\pi}2\sum_{k\geqslant N+1}k^{-3/2},$$ which gives that for each integer $N$, $$\limsup_{n\to +\infty}\left|\sum_{k=1}^{+\infty}a(n,k)\right|\leqslant\sum_{k\geqslant N+1}k^{-3/2}.$$ Conclude (we actually used dominated convergence theorem).

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Can you please explain how you got the line after "which gives that for" and where did you use LDCT ? This means that the sum is $0$, right ? – Belgi Apr 1 '13 at 13:04
Taking the $\limsup$ on both sides, the first sum converges to $0$ because we considered finitely many terms which converge to $0$. We took bound which doesn't depend on $n$. – Davide Giraudo Apr 1 '13 at 13:08
@Davide Giraudo It was not to calculate the value of the serie? – MathOverview Apr 1 '13 at 13:08
@E.Costa It seems we just need the limit as $n\to +\infty$. – Davide Giraudo Apr 1 '13 at 13:09
@Belgi I don't think so... for each $k$, we have that $(-1)^{k}\frac{\arctan{n^{2}k}}{n^{\alpha}+k^{3/2}}\to 0$ as $n\to\infty$, because the numerator tends to $\pi/2$ for $k>0$ and in the denominator we have $k^{3/2}$, which does not change, and $n^\alpha$, which tends to $\infty$ for $\alpha>0$... – Johnny Westerling Apr 1 '13 at 16:55