2
$\begingroup$

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 $n=1$.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ (+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
  • 1
    $\begingroup$ Here is a simple, pre-calculus proof, that $\log(x)\le \sqrt x$. $\endgroup$ – Mark Viola Mar 15 '18 at 18:07
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\frac{e^t}{\sqrt{e^t}}=e^{t/2}$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.