Flawed proof that $\sum_{n=3}^\infty \frac{1}{n\ln{n}(\ln{(\ln{n})})}$ converges.

Question

My proof:

If $\ n \ge 25$, then $$ \frac{1}{n\ln{n}(\ln{(\ln{n})})} \le \frac{1}{n(\ln{n})^{1.1}}$$.

Since $$\sum_{n=2}^\infty \ \frac{1}{n(\ln{n})^{p}}$$ converges when $p \gt 1$.

It follows that $$\sum_{n=3}^\infty \frac{1}{n\ln{n}(\ln{(\ln{n})})}$$ also converges.

I know this series is supposed to diverge and therefore my proof is flawed by I can't see where exactly I'm going wrong.

I think it's the first inequality that is the problem given that I obtained it using Mathematica and didn't actually proof it.

Answer 1

You've forgotten to invert the direction of inequality in your proof. It is not the case that if $a\leq b$ then $\frac1a\leq\frac1b$; in fact, (assuming all numbers involved are positive) the opposite holds: $a\leq b\implies \frac1a\geq\frac1b$. In this case, since $\ln(\ln n)\leq (\ln n)^{0.1}$, we have $\displaystyle\frac1{\ln\ln n}\geq \frac{1}{(\ln n)^{0.1}}$ and thus $\displaystyle\frac1{n(\ln n)(\ln\ln n)}\geq\frac1{n(\ln n)^{1.1}}$. But being termwise greater than the terms of a convergent series tells us nothing about the convergence of the series we're interested in.

Answer 2

Using logarithms base $e,$ what is the derivative of $$ f(x) = \log \log \log x \; \; ? $$ In case of doubt, this means $$ f(x) = \log \left( \log \left( \log x \right) \right) $$

Answer 3

Note that by Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

thus

$$ \sum_{n=1}^{\infty} \frac{2}{n\ln{n}\cdot(\ln{(\ln{n})})} \ge \sum_{n=1}^{\infty} \frac{2^n}{2^n\cdot\ln2^n \cdot\ln (\ln 2^n)}= \sum_{n=1}^{\infty} \frac{1}{n\ln 2\cdot\ln(n\ln 2)}=\sum_{n=1}^{\infty} \frac1{n\ln 2(\ln n+\ln\ln2)}=+\infty$$

indeed

$$\sum_{n=1}^{\infty} \frac2{n\ln 2(\ln n+\ln\ln2)}\ge\sum_{n=1}^{\infty} \frac{2^n}{2^n\ln 2(\ln2^n+\ln\ln2)}=\sum_{n=1}^{\infty} \frac{1}{n\ln^22+\ln2\cdot\ln\ln2}=+\infty$$

Answer 4

If n≥25, then

$\frac{1}{n\ln{n}(\ln{(\ln{n})})} \le \frac{1}{n(\ln{n})^{1.1}}$

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this estimation might not be sharp enough. that could be the flaw in your proof. @will gagy provided elegant solution, which is integral test.