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.

1. \(\displaystyle\sum_{n=2}^\infty\frac{1}{(\log n)^p}\) for any real \(p\)
2. \(\displaystyle\sum_{n=2}^\infty\frac{1}{(\log n)^n}\)
3. \(\displaystyle\sum_{n=2}^\infty\frac{1}{(\log n)^{\log n}}\)
4. \(\displaystyle\sum_{n=3}^\infty\frac{1}{(\log n)^{\log\log n}}\)
5. \(\displaystyle\sum_{n=2}^\infty\frac{1}{n \log n}\)
6. \(\displaystyle\sum_{n=3}^\infty\frac{1}{n (\log n) (\log\log n)^2}\)
7. \(\displaystyle\sum_{n=2}^\infty\frac{\log n}{e^{\sqrt{n}}}\)
8. \(\displaystyle\sum_{n=0}^\infty\frac{n!}{e^{n^2}}\)
9. \(\displaystyle\sum_{n=1}^\infty\log\left(1+\frac{1}{\sqrt{n}}\right)\)
10. \(\displaystyle\sum_{n=1}^\infty(\sqrt[n]{n}-1)^n\)
11. \(\displaystyle\sum_{n=1}^\infty(\sqrt[n]{n}-1)\)

Answers/Hints

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

Very Big Hints

1. Obvious for \(p \le 0\). For \(p < 0\), we have \(\log n < n^{1/p}\) for sufficiently large \(n\).
2. \((\log n)^n > 2^n\) for sufficiently large \(n\).
3. \((\log n)^{\log n} = e^{(\log n)(\log \log n)} = n^{\log\log n} > n^2\) for sufficiently large \(n\).
4. \((\log n)^{\log\log n} = e^{(\log\log n)^2} < e^{(\sqrt{\log n})^2} = n\) for sufficiently large \(n\).
5. Integrate by taking \(u = \log x\).
6. Integrate by taking \(u = \log\log x\).
7. For sufficiently large \(n\) we have \(\log n < n^{1/4}\) and \(e^{\sqrt{n}} > e^{(\log n)^2} = n^{\log n} > n^2\).
8. \(n!/e^{n^2} < n^n/e^{n^2} = (n/e^n)^n < (1/2)^n\) for sufficiently large \(n\).
9. \(\log(1+1/\sqrt{n}) \ge \log(1+1/n) = \log(n+1) – \log n\), and \(\sum_{n=1}^N \log(1+1/n) = \log(N+1)\).
10. 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.
11. 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.