Here are some series convergence problems that I gathered quite a while ago. A few of them are a bit tricky.

Determine convergence/divergence for the nonnegative series below. Prove your answers.

- \(\displaystyle\sum_{n=2}^\infty\frac{1}{(\log n)^p}\) for any real \(p\)
- \(\displaystyle\sum_{n=2}^\infty\frac{1}{(\log n)^n}\)
- \(\displaystyle\sum_{n=2}^\infty\frac{1}{(\log n)^{\log n}}\)
- \(\displaystyle\sum_{n=3}^\infty\frac{1}{(\log n)^{\log\log n}}\)
- \(\displaystyle\sum_{n=2}^\infty\frac{1}{n \log n}\)
- \(\displaystyle\sum_{n=3}^\infty\frac{1}{n (\log n) (\log\log n)^2}\)
- \(\displaystyle\sum_{n=2}^\infty\frac{\log n}{e^{\sqrt{n}}}\)
- \(\displaystyle\sum_{n=0}^\infty\frac{n!}{e^{n^2}}\)
- \(\displaystyle\sum_{n=1}^\infty\log\left(1+\frac{1}{\sqrt{n}}\right)\)
- \(\displaystyle\sum_{n=1}^\infty(\sqrt[n]{n}-1)^n\)
- \(\displaystyle\sum_{n=1}^\infty(\sqrt[n]{n}-1)\)

## Answers/Hints

- No, compare with harmonic series.
- Yes, compare with geometric series or use ratio/root test.
- Yes, compare with \(\sum 1/n^2\).
- No, compare with harmonic series.
- No, use integral test.
- Yes, use integral test.
- Yes, compare with \(\sum 1/n^{3/2}\).
- Yes, compare with geometric series.
- No, compare with \(\log(1+1/n)\) and consider partial sums.
- Yes, use root test.
- No, compare with harmonic series. It may help to show that \((1+1/n)^n \le 3\) for all \(n\).

## Very Big Hints

- Obvious for \(p \le 0\). For \(p < 0\), we have \(\log n < n^{1/p}\) for sufficiently large \(n\).
- \((\log n)^n > 2^n\) for sufficiently large \(n\).
- \((\log n)^{\log n} = e^{(\log n)(\log \log n)} = n^{\log\log n} > n^2\) for sufficiently large \(n\).
- \((\log n)^{\log\log n} = e^{(\log\log n)^2} < e^{(\sqrt{\log n})^2} = n\) for sufficiently large \(n\).
- Integrate by taking \(u = \log x\).
- Integrate by taking \(u = \log\log x\).
- For sufficiently large \(n\) we have \(\log n < n^{1/4}\) and \(e^{\sqrt{n}} > e^{(\log n)^2} = n^{\log n} > n^2\).
- \(n!/e^{n^2} < n^n/e^{n^2} = (n/e^n)^n < (1/2)^n\) for sufficiently large \(n\).
- \(\log(1+1/\sqrt{n}) \ge \log(1+1/n) = \log(n+1) – \log n\), and \(\sum_{n=1}^N \log(1+1/n) = \log(N+1)\).
- Use the root test since \(\sqrt[n]{n}-1 \rightarrow 0\). Alternatively, for \(n \ge 2\) we have $$

n = (1+(\sqrt[n]{n}-1))^n \ge \frac{n(n-1)}{2} (\sqrt[n]{n}-1)^2

$$ by the binomial theorem, so $$

\sqrt[n]{n}-1 \le \sqrt{\frac{2}{n-1}}.

$$ Now $$

(\sqrt[n]{n}-1)^n \le \left(\frac{2}{n-1}\right)^{n/2},

$$ and the comparison test can be applied. - For all \(n\) we have $$

\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^n \left(\frac{1}{k!}\right) \frac{n(n-1) \cdots (n-k+1)}{n^k} \le \sum_{k=0}^n \frac{1}{k!} \le 3,

$$ so \((1+1/n)^n \le 3 \le n\) for all \(n \ge 3\). Therefore \(1+1/n \le \sqrt[n]{n}\), and the comparison test can be applied.

Good work! Here is a question about problem 9. Why the equation in the alternative method is (n^(1/n)-1)^n <= 2^(1/2)/*** but not (n^(1/n)-1)^n <= 2^(n/2)/***

I mean the inequality in problem 10… Sorry for the mistake…

Thank you, I have corrected the article.