I was asked to show that the p-series $\sum_{n=1}^{\infty} 1/n^p$ is convergent without using the integral and limit test. Also, I need to show that $$\sum_{n=1}^{2^k-1} \frac{1}{n^p} < \sum_{n=0}^{k-1} \left(\frac{1}{2^ { p-1}}\right)^n$$ is one of the intermediate step. How can I derive this intermediate step ( using sub-sequence I think) and the following.
*I have google and also search here for helps, and I do see some similar question and good answers, but I still don't know how to derive that intermediate step and the rest.
Any help will be appreciated and thanks in advance.
The intermediate step inequality can be used to show that the $p$-series $\sum_{n=1}^{\infty} 1/n^p$ is convergent for $p>1$ by recalling that the geometric series $\sum_{n=0}^{\infty}x^n$ is convergent when $|x|<1$.
Hint for the intermediate step: for $k>1$, $$\sum_{n=1}^{2^k-1} \frac{1}{n^p}=\sum_{m=0}^{k-1} \sum_{n=2^{m}}^{2^{m+1}-1} \frac{1}{n^p}< \sum_{m=0}^{k-1} \sum_{n=2^{m}}^{2^{m+1}-1} \frac{1}{(2^m)^p}.$$