Considering $0Problem: Considering $0<s<1$ for the series $\sum_{i=1}^{\infty} \frac {1}{i^s}$, I want to show that $a_{2^{n+1} -1}$ $<\sum_{j=0}^{n}$ $({\frac {1} {2^{s-1}})}^j$ where $a_n$ is the partial sums of the aforementioned series..
So for $n=1$ this would yield the inequality:
$\frac {1}{1^s} +\frac {1}{2^s} + \frac {1}{3^s}+ \frac {1}{4^s} < (\frac {1} {2^{s-1}})^{0} + (\frac {1} {2^{s-1}})^{1}=1+\frac {1} {2^{s-1}}$ which would imply that
$\frac {1} {2^{s-1}}  > \frac {1}{2^s} + \frac {1}{3^s} + \frac {1}{4^s} $
which I think I can show since $ \frac {1}{2^s} + \frac {1}{3^s} + \frac {1}{4^s} < \frac {3} {2^{s-1}} < \frac {2} {2^{s-1}}= \frac {1} {2^s} < \frac {1} {2^{s-1}}$.
I just need a way to generalize this notion of grouping the terms in groups of $log_2$ to complete the desired statement. Any hints/help appreciated.
Edit: It seems as if I have mis-written the question several times. I will post it from the notes I have and if it is deemed an invalid question, well then I guess we have our answer. We are answering part (b)
 A: $$
\begin{align}
a_{2^{n+1}} &= \sum_{i=1}^{2^{n+1}} \frac 1{i^s} \\
&= \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{i^s} \\
& \leq \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{2^{js}} \\
& = \sum_{j=0}^{n} \frac {2^j}{2^{js}} \\
& = \sum_{j=0}^{n} \left(\frac 1{2^{s-1}}\right)^j \\
\end{align}
$$
A: $$ \begin{align} 
{\large\sum_{\normalsize i=1}^{\normalsize 2^{n+1}}}\frac{1}{i^s} &= \left[\frac{1}{1^s}\right]+\left[\frac{1}{2^s}+\frac{1}{3^s}\right]+\left[\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}+\frac{1}{7^s}\right]+\cdots+\left[\frac{}{}\cdots\frac{}{}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm] 
&\lt \left[\frac{1}{1^s}\right]+\left[\frac{1}{2^s}+\frac{1}{2^s}\right]+\left[\frac{1}{4^s}+\frac{1}{4^s}+\frac{1}{4^s}+\frac{1}{4^s}\right]+\cdots+\left[\frac{}{}\cdots\frac{}{}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm] 
&= \left[\frac{1}{1^s}\right]+\quad\left[\frac{2}{2^s}\right]\quad+\qquad\qquad\left[\frac{4}{4^s}\right]\qquad\qquad+\cdots+\left[\frac{2^n}{\left(2^n\right)^s}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm] 
&= \frac{1}{\left(2^0\right)^{s-1}}+\frac{1}{\left(2^1\right)^{s-1}}+\frac{1}{\left(2^2\right)^{s-1}}+\cdots+\frac{1}{\left(2^n\right)^{s-1}}\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm] 
&={\large\sum_{\normalsize j=0}^{\normalsize n}}\left(\frac{1}{2^{s-1}}\right)^j\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} 
\end{align} $$ 
Hence,
$$ \boxed{ \,\\ \quad {\Huge s_{{\normalsize 2^{n+1}\color{red}{-1}}}}={\large\sum_{\normalsize i=1}^{\normalsize 2^{n+1}\color{red}{-1}}}\frac{1}{i^s} \quad{\Large\lt}\quad {\large\sum_{\normalsize j=0}^{\normalsize n}}\left(\frac{1}{2^{s-1}}\right)^j \qquad\colon\quad s\,\gt\,0 \quad \\\, } $$ 
   
NB: Part of this answer is a replication of an existing answer by @B.Mehta with corrections.
