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.


[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

Let $f(s)$ be defined as your sum. Let $g(s)=\sum\frac{(-1)^{n+1}}{n^{ns}}$ be the alternating version.


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


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


which finally gives


Lower bounds may likewise be obtained by noticing that


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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