I'm trying to understand whether the following series converges absolutely, conditionally or diverges.

$$\sum _{n=1}^{\infty }\left(-1\right)^{n+1}\left(1-cos\left(\frac{1}{\sqrt{n}}\right)\right)$$

I think that it converges conditionally. But it's getting complicated for me to show that it does not converge absolutely. I've already tried the limit comparison test with $b_n = \frac{1}{\sqrt{n}}$ but the limit results in $0$. I've also tried the integral test, as the conditions for the test hold, but the integral became very complicated.

Is there an easier way?

  • 1
    $\begingroup$ Try $b_n = \frac 1n$. $\endgroup$ – Gabriel Romon May 26 '18 at 14:59
  • $\begingroup$ Thank you, worked out fine with $\frac{1}{n}$. $\endgroup$ – Alon Weissfeld May 26 '18 at 20:12
  • $\begingroup$ @AlonWeissfeld Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Jun 22 '18 at 20:47

Note that

$$1-cos\left(\frac{1}{\sqrt{n}}\right)= \frac1{2n}+O\left(\frac1{n\sqrt n}\right)$$

then the given series doesn’t converge absolutely by limit comparison test with $\sum \frac1{n}$ and then

$$\sum _{n=1}^{\infty }\left(-1\right)^{n+1}\left(1-cos\left(\frac{1}{\sqrt{n}}\right)\right)=\sum _{n=1}^{\infty }\left(-1\right)^{n+1} \frac1{2n}+\sum _{n=1}^{\infty }\left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right)$$

which converges since

  • $\sum _{n=1}^{\infty }\left(-1\right)^{n+1} \frac1{2n}$ converges by alternating series test

  • $\sum _{n=1}^{\infty }\left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right)$ converges absolutely by limit comparison test with $\sum \frac1{n\sqrt n}$

  • $\begingroup$ The last series converges absolutely by the comparison test, not limit comparison. $\endgroup$ – Gabriel Romon May 27 '18 at 5:53
  • $\begingroup$ @GabrielRomon Why limit comparison test shouldn't work? $\endgroup$ – gimusi May 27 '18 at 6:12
  • $\begingroup$ $$\left| \left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right) \right| \leq M \frac1{n\sqrt n}$$ this doesn't say anything else, so you cannot apply limit comparison test. Just the usual comparison test. $\endgroup$ – Gabriel Romon May 27 '18 at 6:45
  • $\begingroup$ @GabrielRomon Why this $$\frac{\left| \left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right) \right|}{ \frac1{n\sqrt n}} \to L \leq M$$ doesn't work? $\endgroup$ – gimusi May 27 '18 at 6:50
  • $\begingroup$ You have absolutely no way to know whether the limit $\lim_n \frac{\left| \left(-1\right)^{n+1} O\left(\frac1{n\sqrt n}\right) \right|}{ \frac1{n\sqrt n}}$ exists ! Just because $u_n\leq M v_n$ doesn't imply that $\lim_n \frac{u_n}{v_n}$ exists. All you can say is $\limsup_n \frac{u_n}{v_n} \leq M$, but that's enough to prove convergence of the series. $\endgroup$ – Gabriel Romon May 27 '18 at 6:56

Using the alternating series test, we see that $\sum_{n=1}^\infty (-1)^{n+1}\left( 1-\cos\left(\frac{1}{\sqrt{n}}\right)\right)$ converges.

From Taylor series we have $$\cos\left(\frac{1}{\sqrt{n}}\right) \leq 1-\frac{\left(\frac{1}{\sqrt{n}}\right)^2}{2} + \frac{\left(\frac{1}{\sqrt{n}}\right)^4}{24}= 1-\frac{1}{2n} + \frac{1}{24n^2}\quad \text{for $n\geq 1$}$$ Therefore, $$\sum_{n=1}^\infty \left(1-\cos\left(\frac{1}{\sqrt{n}}\right) \right)\geq \sum_{n=1}^\infty \left(\frac{1}{2n} -\frac{1}{24n^2}\right)= \infty$$ So, the series fails to converge absolutely.


I think the integral test works.

The taylor series of $\cos(x)$ is $$\sum^\infty_{k=0}\frac{(-1)^k}{(2k)!}x^{2k}$$

Thus, $$1-\cos(\frac1{\sqrt{n}})= -\sum^\infty_{k=1} \frac{(-1)^k}{(2k)!}\frac1{n^{k}}$$

$$-\int^\infty_1 \sum^\infty_{k=1} \frac{(-1)^k}{(2k)!}\frac1{x^{k}}dx=-\int^\infty_1 \sum^\infty_{k=2} \frac{(-1)^k}{(2k)!}\frac1{x^{k}}dx+\int^\infty_1\frac{1}{2x}dx=C+\infty$$ where $C$ is finite.

Therefore the sum does not converge absolutely.

It is an elementary exercise to prove $C$ is finite.:)

  • $\begingroup$ After the first line, you assume taylor series is allowed? When I took Calculus II, taylor series was after convergence... $\endgroup$ – BCLC May 26 '18 at 15:23
  • $\begingroup$ @BCLC I am not clear about the mathematics syllabus of foreign countries. $\endgroup$ – Szeto May 27 '18 at 4:13
  • $\begingroup$ you seriously learned Taylor series first? In Stewart calculus Taylor series comes after right? $\endgroup$ – BCLC May 27 '18 at 6:53
  • $\begingroup$ @BCLC Well, I learned things before the school taught me. I usually learn those stuff from MathSOS, here MathSE, or some documents showed up in Google search. I don’t really remember which should be taught first. Moreover, for some common entire functions, a student can use its taylor series directly. Using taylor series is easy when one does not have to understand it and care about its convergence. $\endgroup$ – Szeto May 27 '18 at 6:57

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