Convergence/Divergence of the series $\sum_{n=1}^\infty\frac{(-1)^n\log n}{\sqrt n}$

Question

According to the answers in my book, the series converges.

I know I should use Alternating Series Test and the limit of ln (n)/(sqrt n) approaches 0.

And the derivative is negative after n=8, but the derivative is positive when 1<=n<=8, is that ok to use Alternating Series Test? I remember the condition is that it should decrease for all n>=1.

Answer 1

It seems to be you mean the series $\;\displaystyle\sum\limits_{n=1}^\infty\frac{(-1)^n\log n}{\sqrt n}\;$ . Now, we have

$$f(x):=\frac{\log x}{\sqrt x}\implies f'(x)=\frac{\frac1{\sqrt x}-\frac{\log x}{2\sqrt x}}{x}=\frac{2-\log x}{2x\sqrt x}<0\iff\log x>2\iff x>e^2$$

and then the sequence $\;\left\{\,\frac{\log n}{\sqrt n}\,\right\}_{n\in\Bbb N}\;$ is mononotonic decreasing ( and clearly positive for all $\;n>1\;$) for $\;n>e^2\implies n\ge8\;$ , and thus we have here a Leibniz series which converges (conditionally, though)

Answer 2

Not an answer, just a note.

We can actually find the closed form of this series, by considering a more general one:

$$f(s)=\sum_{n=1}^\infty (-1)^n \frac{\log n}{n^s}$$

Let's introduce an auxiliary function:

$$g(s)=\sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^s}=(1-2^{1-s}) \zeta(s)$$

This is true by definition of http://mathworld.wolfram.com/DirichletEtaFunction.html. Though, we can prove it independently as well.

On the other hand, we can see:

$$f(s)=\frac{d}{ds} g(s)$$

So we can immediately write:

$$f(s)=\frac{d}{ds} \left[(1-2^{1-s}) \zeta(s) \right]=2^{1-s} \zeta(s) \log 2 +(1-2^{1-s}) \zeta'(s)$$

Derivative of zeta function is usually quite complicated and doesn't always have a closed form.

However for $s=1/2$ it's a special case which allows us to use the so called "reflection formula" to get the closed form, as can be seen from this answer: https://math.stackexchange.com/a/2495423/269624

$$\zeta'\left(\frac{1}{2}\right) = \frac{1}{2} \zeta\left(\frac{1}{2}\right) \left( \log \pi+\gamma+\frac{\pi}{2}+3\log 2 \right)$$

Which finally gives us:

$$f \left(\frac{1}{2}\right) =\frac{1}{2} \zeta\left(\frac{1}{2}\right) \left[2\sqrt{2} \log 2-(\sqrt{2}-1) \left( \log \pi+\gamma+\frac{\pi}{2}+3\log 2 \right) \right]$$