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Determine whether the following series converges conditionally, or converges absolutely. $$\sum^\infty_{k=2}\frac{\sin(\frac{\pi}{2}+k\pi)}{\sqrt{k}\ln(k)}$$

What could I use here to work this out? This isn't monotone. I've tried using the Ratio Test but this seems kinda cumbersome. Is there a better way to do this?

Was I even allowed to use the Ratio Test? I've just remembered it's not for $\sum^\infty_{n=c}$ but for $n=1$.

AustereTiger
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  • Regarding your last question, in issues of whether things converge or not, where you start the series is always irrelevant. Convergence is about the infinite tail of the series – spaceisdarkgreen Mar 15 '18 at 14:58

3 Answers3

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Note that

$$\sum^\infty_{k=2}\frac{\sin(\frac{\pi}{2}+k\pi)}{\sqrt{k}\ln(k)}=\sum^\infty_{k=2}\frac{(-1)^k}{\sqrt{k}\ln(k)}$$

thus it converges conditionally by alternating series test while

$$\sum^\infty_{k=2}\left|\frac{\sin(\frac{\pi}{2}+k\pi)}{\sqrt{k}\ln(k)}\right|=\sum^\infty_{k=2}\frac{1}{\sqrt{k}\ln(k)}$$

diverges by limit comparison test with $\frac{1}{k^\frac34}$.

For the latter, as an alternative suggested by Mark Viola, note that since for any $a>0$

$$\log x^a\le x^a-1 \implies\log x\le \frac{x^a-1}{a}<\frac{x^a}{a}$$

selecting $a=\frac12$ we have

$$\sum^\infty_{k=2}\frac{1}{\sqrt{k}\ln(k)}>\sum^\infty_{k=2}\frac{1}{2k}$$

Mark Viola
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user
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    @ClementC. Oh yes of course, Thanks I fix – user Mar 15 '18 at 15:01
  • Note that for any $a>0$, we have $$\log(x)\le \frac{x^a-1}{a}<\frac{x^a}{a}$$So, taking any $0<a\le 1/2$ suffices here. – Mark Viola Mar 15 '18 at 15:25
  • @MarkViola Yes Mark of course I've just take a value in this range (doesn't it should be 0<a< 1/2?). – user Mar 15 '18 at 15:45
  • Since the harmonic series diverges, we can select $a=1/2$ here. – Mark Viola Mar 15 '18 at 15:46
  • @MarkViola I'm referring to limit comparison test with "a" such that $$\frac{\frac{1}{\sqrt{k}\ln(k)}}{\frac1{k^a}}\to \infty$$ and with "a=1/2" I could not conclude. – user Mar 15 '18 at 15:50
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    Yes, but I'm suggesting you forgo that and use the comparison test, which I believe is more straightforward and easier to understand intuitively than the LCT. – Mark Viola Mar 15 '18 at 16:11
  • @MarkViola Oh yes very nice Mark, I'm addicted with LCT but in this case that really more straightforward and clear. – user Mar 15 '18 at 16:38
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    (+1) for the edited result. I corrected for the missing factor of $2$ in the denominator. As it turns out, one can show with more work that $\log(k)\le \sqrt k$ for $k\ge 2$. – Mark Viola Mar 15 '18 at 17:25
  • @MarkViola It's an upovote for you and thanks for the edit, I had considered 2/k but 1/(2k) is the correct bound! – user Mar 15 '18 at 17:29
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    Here is a simple, pre-calculus proof, that $\log(x)\le \sqrt x$. – Mark Viola Mar 15 '18 at 18:07
  • @MarkViola Thanks a lot Mark I will study you work on it! – user Mar 15 '18 at 18:10
  • I get to showing the limit of $a_k$ is zero (as required) and $\frac{1}{\infty}$ shows up. Would this just be zero? – AustereTiger Mar 15 '18 at 18:47
  • @AustereTiger are you referring to the alternating series test or to the second one? – user Mar 15 '18 at 18:51
  • @gimusi The alternating series test. – AustereTiger Mar 15 '18 at 18:54
  • @gimusi Using calculus it's easy to show that $\log(x)\le x^a$ for all $x>0$ whenever $a\ge 1/e$. – Mark Viola Mar 15 '18 at 18:56
  • @AusterTiger $a_k \to 0$ is always necessary. To apply alternating series test to $\sum (-1)^k a_k$ we need that $a_k$ is monolitically decreasing which is the case here since $\sqrt k \ln k$ is the product of two increasing functions thus the reciprocal is strictly decreasing. – user Mar 15 '18 at 19:03
  • @gimusi That's what I'm saying. $a_k$ must be approaching zero. To apply the alternating series test in the first place, surely I need to show that the limit is actually zero? – AustereTiger Mar 15 '18 at 19:05
  • @AustereTiger Yes of course that’s a necessary condition for al kind of series to converge. – user Mar 15 '18 at 19:08
  • That's what my problem is then. As I said, $\frac{1}{\infty}$ shows up in the limit calculation. That's one of those unintuitive ones where you could probably just say it equals zero but I would rather show it. I was thinking of using the Squeeze Theorem to show the limit is zero. – AustereTiger Mar 15 '18 at 19:09
  • @AustereTiger We don’t need squeeze theorem since it is not an indeterminate form you can refer to it as a given result by limit theorems http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx – user Mar 15 '18 at 19:18
  • @gimusi Thank you for that :) I'll look at the proof. – AustereTiger Mar 15 '18 at 19:21
  • @AustereTiger You are welcome, please do not hesitate to ask for any other clarification, I don't know your level of knowledge on this topic thus maybe I've assumed mny things as given but if something is not clear you can ask for it. Bye – user Mar 15 '18 at 19:26
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Note that $\sin(\pi/2 +k\pi)$ just alternates between $1$ and $-1,$ so has absolute value $1$. So absolute convergence is just about $\frac{1}{\sqrt k \ln k}.$ For this I recommend a comparison with $1/k$. For the convergence of the series itself I recommend the alternating series test.

spaceisdarkgreen
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Firstly, we will study if the serie $$\sum_{k=2}^\infty\left|\frac{\sin\left(\frac{\pi}{2}+k\pi\right)}{\sqrt{k}\ln(k)}\right|$$ converges, because if it does, directly we have that our original one does too. Note that $$\sin\left(\frac{\pi}{2}+k\pi\right)=(-1)^{k}$$ so we have that $$\sum_{k=2}^\infty\left|\frac{\sin\left(\frac{\pi}{2}+k\pi\right)}{\sqrt{k}\ln(k)}\right|=\sum_{k=2}^\infty\left|\frac{1}{\sqrt{k}\ln(k)}\right|=\sum_{k=2}^\infty\frac{1}{\sqrt{k}\ln(k)}$$

Now, we're going to use the integral test for convergence

https://en.wikipedia.org/wiki/Integral_test_for_convergence

As the integral $$\int \frac{1}{\sqrt{x}\ln(x)}dx=\int_{\ln(2)}^\infty \frac{e^t}{t\sqrt{e^t}}dt =\int_{\ln(2)}^\infty \frac{1}{t\sqrt{e^{-t}}}dt $$ has not a primitive on elementary function, we can proceed comparing this integral with another and apply the comparison test for improper integrals

http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

Comparing it with $g(t)=\frac{1}{t}$, we get that $$\lim_{t\to \infty} \frac{\frac{1}{t\sqrt{e^{-t}}}}{\frac{1}{t}}=\lim_{t\to \infty} \frac{1}{\sqrt{e^{-t}}}=+\infty$$ and as $g$ diverges, $f$ does it too.

user326159
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  • $\frac{e^t}{\sqrt{e^t}}=e^{t/2}$ and $\int_1^L \frac{e^{t/2}}{t},dt\ge \int_1^L\frac1t,dt$, which diverges logarithmically. But we don't need the integral test here. Note that $\log(k)\le \frac{k^a-1}{a}$ for all $a>0$. So, just choose $a$ to equal anything less than or equal to $1/2$ and you're done since the harmonic series diverges. – Mark Viola Mar 15 '18 at 15:22