Valid Series Convergence Proof?

Question

Just wanted someone to check me to see if this is a valid proof. I've seen other proofs, but I can't use Cauchy condensation, or integral test as we haven't learned it yet. I have ratio test and comparison test at my disposal.

Show how that $\displaystyle\sum_{n=2}^\infty\frac1{n\log n}$ diverges.

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$n>0$, so for all $n$, $0<(n \log n)^{-1}$. By $\Bbb Q$ dense in $\Bbb R$, for all $n$ there exists a $q$ such that

$$0<q<\frac1{n\log n}<\frac1{2\log2}$$

By the Archimedean property, there exits an $m$ such that $mq>(2\log2)^{-1}$. Therefore,

$$0<\frac1{2m\log2}<q<\frac1{n\log n}<\frac1{2\log2}$$

As $\sum 1/m$ diverges as it is the tail harmonic series, $1/(n\log n)$ diverges.

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I'm a bit unsure of this, but I have no idea how else to show it using the tools I have.

Answer 1

Note that using the inequality $\log(x)\le x-1$ (See This Answer), it is easy to show

$$\begin{align} \log(\log(n+1))-\log(\log(n))&=\log\left(\frac{\log(n+1)}{\log(n)}\right)\\\\ &\le\frac{\log(n+1)-\log(n)}{\log(n)}\\\\ &\le \frac1{n\log(n)}\tag 1 \end{align}$$

Summing both sides of $(1)$ reveals

$$\sum_{n=2}^N \frac{1}{n\log(n)}\ge \log(\log(N+1))-\log(\log(2))\to \infty$$

as $N\to \infty$. And we are done!

Answer 2

No, your proof is not correct. The reason is as follows: your argument amounts to

for every $n$ there exists $m(n)$ such that $$ \frac{1}{(2\log 2)m(n)} < \frac{1}{n\log n} $$ and therefore $\sum_{m=1}^N \frac{1}{(2\log 2)m} < \sum_{n=1}^N \frac{1}{n\log n}$; but the LHS diverges.

The "therefore" is where things go haywire. You are summing $\frac{1}{m(n)}$ for $n\geq 1$, not $\frac{1}{m}$ for $m\geq 1$. And nothing tells you that the series with partial sums $$\sum_{n=1}^N \frac{1}{m(n)}$$ diverges. (For instance, why couldn't it be for instance the case that $m(n) = 2^n$)?

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Now, as you mention two usual ways to prove divergence of $\sum_n \frac{1}{n\log n}$ are either the Cauchy condensation criterion, or the comparison series-integral (noting that a primitive of $\frac{1}{x\log x}$ is $\log\log x$). You may not use them yet, but once you learn them I encourage you to try to apply them to this case. (The ratio test will not help you here, and neither the comparison test -- for instance, with $p$-series.)