# Determining if a series converges conditionally or absolutely

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 $n=1$.

• 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

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

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

• @ClementC. Oh yes of course, Thanks I fix – gimusi 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
• 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
• (+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
• Here is a simple, pre-calculus proof, that $\log(x)\le \sqrt x$. – 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.

• But for the alternating series test, doesn't the series have to be non-increasing? – AustereTiger Mar 15 '18 at 14:55
• Yes, and it is. – spaceisdarkgreen Mar 15 '18 at 14:56
• It is? How do we know? – AustereTiger Mar 15 '18 at 14:59
• 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 $(1,\infty)$). – Clement C. Mar 15 '18 at 15:01
• 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. – 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

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.

• $\frac{e^t}{\sqrt{e^t}}=e^{t/2}$ 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