TITLE: Convergence/Divergence of some series
QUESTION [8 upvotes]: If I have $\sum_{n=2}^{\infty} \frac {1}{n\log n}$ and want to prove that it diverges, can I use following?
$$\frac {1}{n\log n} \lt \frac {1}{n}$$

$\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of $\frac 1 n$ equals to zero so the comparasion I think isn't useful.
Or can I say it diverges because $\sum_{n=1}^\infty \frac 1{\log n}$ also diverges?

And If I have $\sum_{n=2}^\infty \frac {(-1)^n}{(n)}$ I can use the Leibniz rule. $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that $\sum_{n=1}^\infty \frac 1 n$ has an infinite sum, but what to do in this case?

REPLY [2 votes]: The really long explanation of what robjohn's answer and Gareth McCaughan's answer are trying to say for those who like it written out.

Perhaps it will help to see what the Cauchy condensation test says:
$$f(n)\ge f(n+k)\implies\\\sum_{n=1}^\infty2^nf(2^{n+1})\ge\sum_{n=2}^\infty f(n)\ge\sum_{n=1}^\infty2^{n-1}f(2^n)$$
Here, we have $f(n)=\frac1{n\log(n)}$.  A visualization of what's happening:
$$\small\begin{align}\sum_{n=2}^\infty\frac1{n\log(n)}&=\frac1{2\log(2)}+\frac1{3\log(3)}+\frac1{4\log(4)}+\frac1{5\log(5)}+\frac1{6\log(6)}+\frac1{7\log(7)}+\frac1{8\log(8)}+\dots\\&\ge\underbrace{\frac1{2\log(2)}}_{2^0f(2^1)}+\underbrace{\frac1{4\log(4)}+\frac1{4\log(4)}}_{2^1f(2^2)}+\underbrace{\frac1{8\log(8)}+\frac1{8\log(8)}+\frac1{8\log(8)}+\frac1{8\log(8)}}_{2^2f(2^3)}+\dots\\&=\sum_{n=1}^\infty2^{n-1}f(2^n)\end{align}$$
It is then noted that:
$$2^{n-1}f(2^n)=\frac1{2n\log(2)}$$
And as we know that $\sum_{n=1}^\infty\frac1n$ diverges, we can see that
$$\sum_{n=2}^\infty\frac1{n\log(n)}\ge\sum_{n=1}^\infty2^{n-1}f(2^n)=\frac1{2\log(2)}\sum_{n=1}^\infty\frac1n=+\infty\\\implies\sum_{n=2}^\infty\frac1{n\log(n)}=+\infty$$

Also note that your comparison is done backwards.  It is always the case that you can find an upper bound that sums to infinity, but to get results from the comparison test, you need to find a lower bound that sums to inifnity, as we've done here.


As for the Leibniz rule and the alternating harmonic series, notice that:
$$\small\begin{align}\sum_{n=1}^\infty\frac{(-1)^{n+1}}n&=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\\&\le\frac11-\frac12+\frac12-\frac14+\frac14-\frac16+\frac16-\frac18+\dots\\&=1\end{align}$$
$$\small\begin{align}\sum_{n=1}^\infty\frac{(-1)^{n+1}}n&=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\\&\ge\frac11-\frac11+\frac13-\frac13+\frac15-\frac15+\frac17-\frac17+\dots\\&=0\end{align}$$
Thus, we have
$$0\le\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\le1$$
And as $\lim_{n\to\infty}\frac{(-1)^{n+1}}n=0$, it must be the case that the alternating harmonic series converges, as it is bounded and the difference between partial sums tends to zero.

On the other hand, the harmonic series is not bounded above.  There are plenty of approaches to showing this.  Using the Cauchy condensation test from above, we find that
$$f(n)=\frac1n\\\begin{align}\sum_{n=1}^\infty\frac1n&=\sum_{n=1}^\infty f(n)\\&\ge1+\sum_{n=1}^\infty2^{n-1}f(2^n)\\&=1+\sum_{n=1}^\infty\frac{2^{n-1}}{2^n}\\&=1+\sum_{n=1}^\infty\frac12\\&=1+\frac12+\frac12+\frac12+\dots\\&=+\infty\end{align}$$