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 $n=1$.

  • $\begingroup$ 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 $\endgroup$ – spaceisdarkgreen Mar 15 '18 at 14:58

Note that


thus it converges conditionally by alternating series test while


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


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    $\begingroup$ @ClementC. Oh yes of course, Thanks I fix $\endgroup$ – gimusi Mar 15 '18 at 15:01
  • $\begingroup$ 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. $\endgroup$ – Mark Viola Mar 15 '18 at 15:25
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    $\begingroup$ 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. $\endgroup$ – Mark Viola Mar 15 '18 at 16:11
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    $\begingroup$ (+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 $k\ge 2$. $\endgroup$ – Mark Viola Mar 15 '18 at 17:25
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    $\begingroup$ Here is a simple, pre-calculus proof, that $\log(x)\le \sqrt x$. $\endgroup$ – Mark Viola Mar 15 '18 at 18:07

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.

  • $\begingroup$ But for the alternating series test, doesn't the series have to be non-increasing? $\endgroup$ – AustereTiger Mar 15 '18 at 14:55
  • $\begingroup$ Yes, and it is. $\endgroup$ – spaceisdarkgreen Mar 15 '18 at 14:56
  • $\begingroup$ It is? How do we know? $\endgroup$ – AustereTiger Mar 15 '18 at 14:59
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    $\begingroup$ The function $x\mapsto \sqrt{x}$ is increasing, and so is $x\mapsto \ln x$. So $x\mapsto \sqrt{x}\ln x$ is increasing, and $x\mapsto \frac{1}{\sqrt{x}\ln x}$ is decreasing (on $(1,\infty)$). $\endgroup$ – Clement C. Mar 15 '18 at 15:01
  • $\begingroup$ The absolute value of the series must be non-increasing, that is. (If that was the confusion.) No alternating series is monotonic, so it’d be a pretty useless test if that were the criterion. $\endgroup$ – spaceisdarkgreen Mar 15 '18 at 15:01

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


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


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.

  • $\begingroup$ $\frac{e^t}{\sqrt{e^t}}=e^{t/2}$ 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. $\endgroup$ – Mark Viola Mar 15 '18 at 15:22

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