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At the minute 3 of the video lecture [1], the professor prove for real numbers $s>1$ the well known statement $$\sum_{n=1}^\infty\frac{1}{n^s}\leq \frac{s}{s-1}\tag{1}$$ With this idea I am interested to know how get an upper bound, now for the series $$\sum_{n=1}^\infty \frac{1}{n^{ns}}$$ for real numbers $s>1$. Thus as did the professor I write using the integral test $$\sum_{n=1}^\infty \frac{1}{n^{ns}}=1+\int_1^\infty\frac{dx}{x^{xs}},$$ and since the domain of integration is $x\geq 1$, and it implies $x^{xs}\geq x^s$ for real numbers $s>1$, then one has $\frac{1}{x^{xs}}\leq \frac{1}{x^s}$. And thus using the direct integration and evaluation of the improper integral that calculated the professor one has the upper bound $$\sum_{n=1}^\infty \frac{1}{n^{ns}}=1+\int_1^\infty\frac{dx}{x^{xs}}\leq 1+\int_1^\infty\frac{dx}{x^{s}}=\frac{s}{s-1}$$ for reals $s>1$. But I don't know how do an improvement of such upper bound $\frac{s}{s-1}$.

Question. I believe that it is feasible improve my calculations. What are your calculations to get an improvement of such upper bound in terms of real numbers $s>1$ here $$\sum_{n=1}^\infty \frac{1}{n^{ns}}\leq\text{ upper bound}?$$ Many thanks.

References:

[1] From YouTube, Week6Lecture4: The Riemann Zeta Function and the Riemann Hypothesis, from the official channel Petra Bonfert-Taylor.

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    $\begingroup$ At $s=1$, we get Sophomore's dream $\endgroup$ – Simply Beautiful Art Apr 22 '17 at 12:44
  • $\begingroup$ Many thanks for your help @SimplyBeautifulArt $\endgroup$ – user243301 Apr 22 '17 at 12:46
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    $\begingroup$ What is the actual purpose of such approximation? The series defining $\zeta(s)$ has a pole at $s=1$, hence the behaviour in a neighbourhood helps us in understanding something about the distribution of prime numbers, but $\sum_{n\geq 1}\frac{1}{n^{ns}}$ converges so fast for any $s>0$ that a tight upper bound is straightforward to find, and pretty useless too. $\endgroup$ – Jack D'Aurizio Apr 22 '17 at 13:32
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    $\begingroup$ Another approach would involve noticing that $\frac1{n^{ns}}<\frac1{k^{ns}}$ for $n>k$... $\endgroup$ – Simply Beautiful Art Apr 22 '17 at 13:36
  • $\begingroup$ @JackD'Aurizio was only a curiosity, an exercise without a special purpose. Many thanks for your remarks and also Simple Beautiful Art. $\endgroup$ – user243301 Apr 22 '17 at 14:01
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Let $f(s)$ be defined as your sum. Let $g(s)=\sum\frac{(-1)^{n+1}}{n^{ns}}$ be the alternating version.

$$f(s)-g(s)=\sum_{n=1}^\infty\frac{1+(-1)^n}{n^{ns}}=\sum_{n=1}^\infty\frac2{(2n)^{2ns}}<2^{1-2s}+\sum_{n=2}^\infty\frac1{(2\cdot2)^{2ns}}=2^{1-2s}+\frac{2^{1-4s}}{2^{4s}-1}$$

which holds for $s>0$. It thus follows that

$$f(s)<2^{1-2s}+\frac{2^{1-4s}}{2^{4s}-1}+g(s)$$

And likewise, it is easy to deduce simple bounds thanks to the alternating series remainder, such as...

$$g(s)<1-2^{-2s}+3^{-3s}$$

which finally gives

$$f(s)>1+2^{-2s}+3^{-3s}+\frac{2^{1-4s}}{2^{4s}-1}$$

Lower bounds may likewise be obtained by noticing that

$$\sum_{n=1}^\infty\frac1{(2n)^{2ns}}=\sum_{n=1}^\infty\frac{2^{-ns}}{n^{2ns}}>2^{-s}f(2s)>2^{-s}f(s)$$

Or by removing the $\frac{2^{1-4s}}{2^{4s}-1}$ term altogether.

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  • $\begingroup$ Many thanks for your answer. $\endgroup$ – user243301 Apr 22 '17 at 13:13
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    $\begingroup$ @user243301 the bound is pretty good: desmos.com/calculator/txbeicfgmw $\endgroup$ – Simply Beautiful Art Apr 22 '17 at 13:15
  • $\begingroup$ I've undertand the first line of your proof, now I am doing coffee. So patience (is a little joke, but truly I've read the first line and I am doing coffee). Many thanks. $\endgroup$ – user243301 Apr 22 '17 at 13:24
  • $\begingroup$ @user243301 :-) well I'm a young one who still drinks milk in the morning. $\endgroup$ – Simply Beautiful Art Apr 22 '17 at 13:28

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