# Some series convergence problems

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

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. $$\displaystyle\sum_{n=1}^\infty(\sqrt[n]{n}-1)$$

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. wj32

1. Mizi says:
1. Mizi says:
2. wj32 says: