Determining if a series converges conditionally or absolutely

Question

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 $n=1$.

Answer 1

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

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

Answer 2

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.

Answer 3

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.