Problem: Considering $0 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)
Considering $0
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0[Cauchy condensation test?](https://en.wikipedia.org/wiki/Cauchy_condensation_test) – 2017-02-21
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1Sorry about the nitpicking, but this may help in preventing downvotes: the first sum should start at $i=1$. – 2017-02-21
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0@HazemOrabi that is incorrect, you are missing the first term on the left, and the second term on the right – 2017-02-24
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0@HazemOrabi you are right, the question was badly worded, will edit. – 2017-02-24
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0In my opinion, it seems the inequality holds for all $\,s\gt\alpha\ge0\,$, where $\,\alpha\,$ is constant depends on $\,n\,$. Can you double check this assumption? – 2017-02-24
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0@HazemOrabi sorry about the confusion I have posted the exact statement from my notes. It is question (b) – 2017-02-24
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0Thanks for the effort. The problem is still exist even if $\,s\gt1\,$. $$ s=6/5,\,n=1\quad\Rightarrow\quad{\large1}+\frac{1}{{\large2}^{\small6/5}}+\frac{1}{{\large3}^{\small6/5}}+\frac{1}{{\large4}^{\small6/5}}\approx1.89232 \,\color{red}{\large\gt}\, {\large1}+\frac{1}{{\large2}^{\small1/5}}\approx1.87055 $$ – 2017-02-24
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0Ok if you post that the question is flawed I will give you the bounty. – 2017-02-24
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0Let us wait and see. I believe something is wrong or missing. – 2017-02-24
2 Answers
$$ \begin{align}
{\large\sum_{\normalsize i=1}^{\normalsize 2^{n+1}}}\frac{1}{i^s} &= \left[\frac{1}{1^s}\right]+\left[\frac{1}{2^s}+\frac{1}{3^s}\right]+\left[\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}+\frac{1}{7^s}\right]+\cdots+\left[\frac{}{}\cdots\frac{}{}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&\lt \left[\frac{1}{1^s}\right]+\left[\frac{1}{2^s}+\frac{1}{2^s}\right]+\left[\frac{1}{4^s}+\frac{1}{4^s}+\frac{1}{4^s}+\frac{1}{4^s}\right]+\cdots+\left[\frac{}{}\cdots\frac{}{}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&= \left[\frac{1}{1^s}\right]+\quad\left[\frac{2}{2^s}\right]\quad+\qquad\qquad\left[\frac{4}{4^s}\right]\qquad\qquad+\cdots+\left[\frac{2^n}{\left(2^n\right)^s}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&= \frac{1}{\left(2^0\right)^{s-1}}+\frac{1}{\left(2^1\right)^{s-1}}+\frac{1}{\left(2^2\right)^{s-1}}+\cdots+\frac{1}{\left(2^n\right)^{s-1}}\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&={\large\sum_{\normalsize j=0}^{\normalsize n}}\left(\frac{1}{2^{s-1}}\right)^j\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}}
\end{align} $$
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.
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0@TheMathNoob: We finally catch it. Thanks. – 2017-02-24
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0Thanks a lot! will review continually throughout the night. – 2017-02-25
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0@TheMathNoob: Thanks. – 2017-03-03
$$ \begin{align} a_{2^{n+1}} &= \sum_{i=1}^{2^{n+1}} \frac 1{i^s} \\ &= \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{i^s} \\ & \leq \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{2^{js}} \\ & = \sum_{j=0}^{n} \frac {2^j}{2^{js}} \\ & = \sum_{j=0}^{n} \left(\frac 1{2^{s-1}}\right)^j \\ \end{align} $$