This is an exercise from Conway. With some help I managed to solve (a), (b) and the first half of (c). However I am stuck with the second inequality of (c). I can't find a way to prove the inequality. Could anyone help me? Also could anyone give me a hint on how to solve (e)? I think I managed to solve (d).
The second inequality of (c):
Note that $$ r^N\le r^{(2r)^{1/(\alpha -\varepsilon )}} $$ and \begin{align} \sum_{n=p+1}^N\left(\frac{1}{n^{\alpha -\varepsilon }}\right)^n&\le \sum_{n=p+1}^\infty\left(\frac{1}{n^{\alpha -\varepsilon }}\right)^n\le\sum_{n=p+1}^\infty\left(\frac{1}{(p+1)^{\alpha -\varepsilon }}\right)^n\\ &\le \left(\frac{1}{(p+1)^{\alpha -\varepsilon }}\right)^{p+1}\sum_{n=0}^\infty\left(\frac{1}{(p+1)^{\alpha -\varepsilon }}\right)^n<\infty. \end{align} Hence we have $$ \sum_{n=p+1}^N\left(\frac{r}{n^{\alpha -\varepsilon }}\right)^n <r^N \sum_{n=p+1}^N\left(\frac{1}{n^{\alpha -\varepsilon }}\right)^n<Br^{(2r)^{1/(\alpha -\varepsilon )}}=B\exp\left( (2r)^{1/(\alpha -\varepsilon )}\log r\right) $$ where $$ B= \left(\frac{1}{(p+1)^{\alpha -\varepsilon }}\right)^{p+1}\sum_{n=0}^\infty\left(\frac{1}{(p+1)^{\alpha -\varepsilon }}\right)^n. $$
Hint for part (e):
In part (a), from the inequality$$ \frac{-\log |c_n|}{n\log n}\ge \frac{1}{\beta }\left(1-\frac{1+\log \beta }{\log n}\right) $$ we can conclude not only $\alpha >0$ but also $\alpha \ge 1/\beta $. Tending $\beta $ to $\lambda $ we have $$ \lambda \ge \frac{1}{\alpha }.$$