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I found this proof: Convergence of Alternating Zeta Series

I do not understand the motivation in going from the first to the second line. What convergence test, if any, have they actually used? Could someone just give more details on how they attempted this proof.

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I think the proof you posted is using the MVT for the function $\;f(z):=\cfrac1{z^s}\;$ on the interval $\;[2n-1,\,2n]\;$ , since then

$$\frac{f(2n)-f(2n-1)}{2n-(2n-1)}=f'(c)\;,\;\;c\in (2n-1,\,2n)$$

But

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

and

$$f'(c)=-\frac s{c^{s+1}}$$

and taking absolute values we get at once

$$|f'(c)|=\left|\frac s{c^{s+1}}\right|\le\frac{|s|}{|2n-1|^{s+1}}\,,\,\,\text{since}\;\;c>2n-1...$$

and the series of the last rightmost expression converges say because $\;\text{Re}\,(s+1)>1\;$ ...

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  • $\begingroup$ I don’t think MVT exists for complex-values functions. That is why I provided a link to a similar inequality (I would be happy to be told I’m wrong, though!). Also, the modulus of a positive real number to a complex power is the number raised to the real part of the complex number. $\endgroup$ – Clayton Feb 15 '18 at 18:00
  • $\begingroup$ @Clayton You're right (although some extensions can be done here and there...), yet I mean the above either for real $\;z\in\Bbb R\;$ or else for the real function $\;|f(z)|=\frac1{|z^s|}\;$ ... $\endgroup$ – DonAntonio Feb 15 '18 at 18:07
  • $\begingroup$ But the modulus isn’t differentiable in general. In the first case ($z\in\Bbb R$), you still have a complex exponent while in the second case, you have a non-differentiable function. I’m not trying to be complicated, I just want to be as clear as I can. $\endgroup$ – Clayton Feb 15 '18 at 18:37
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    $\begingroup$ @Clayton The absolute value (or real modulus, if you prefer) of a differentiable function $\;f(x)\;$ is differentiable everywhere that $\;f(x)\neq0\;$ ... $\endgroup$ – DonAntonio Feb 15 '18 at 18:39
  • $\begingroup$ Thanks, @Don! I didn’t think about where we were in $\Bbb C$, just that the it wasn’t dofferentiable in general. I think that suffices. $\endgroup$ – Clayton Feb 15 '18 at 18:47
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Consider the definition of the derivative: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$$ In the summation, you have something like $$\frac{f(n+1)-f(n)}{1}.$$So all that is happening is that they are making an estimation of the derivative: $$ \left|\frac{f(2n-1)-f(2n)}{1}\right|\leq\max_{x\in[2n-1,2n]}\left\{\left|\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\right\}\leq\left|\frac{s}{(2n-1)^{s+1}}\right|.$$ For a proof of this inequality, see this page.

Once they have this, the resulting sum is absolutely convergent since $s$ is fixed (in particular, $|s|\leq M$) and $|(2n-1)^{s+1}|=(2n-1)^{\mathrm{Re}(s)+1}$. In particular, $$\sum_{n=1}^\infty\frac{|s|}{(2n-1)^{1+\mathrm{Re}(s)}}\leq\sum_{n=1}^\infty\frac{M}{n^{1+\mathrm{Re}(s)}}.$$The latter sum can be shown to converge using the integral test.

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