\(a_{n}=-2\sqrt{n}+\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}\to\zeta(\frac{1}{2})\)

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Grigorios Kostakos
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\(a_{n}=-2\sqrt{n}+\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}\to\zeta(\frac{1}{2})\)

#1

Post by Grigorios Kostakos »

Let \(\{{a_{n}}\}_{n=1}^{\infty}\) a sequence of real numbers such that $$a_{n}=-2\sqrt{n}+\displaystyle\mathop{\sum}\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}\,.$$ a) Prove that the sequence \(\{{a_{n}}\}_{n=1}^{\infty}\) converges.

b) Prove that $$\mathop{\lim}\limits_{n\rightarrow{+\infty}}a_{n}=\zeta\bigl({\tfrac{1}{2}}\bigr)\,,$$ where \({\zeta(s)}\) is (extended) Riemann zeta function.


NOTE: I don't have a solution for b).
Grigorios Kostakos
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Re: \(a_{n}=-2\sqrt{n}+\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}\to\zeta(\frac{1}{2})\)

#2

Post by akotronis »

Hello. Since \(\zeta(1/2)\) cannot be expressed using other known mathematical constants, we pretty much have a matter of definition here. The values of \(\zeta(s)\) for \(\Re(s)\leq1\) , \(s\neq1\) are defined via analytic continuation. This can be done for example using the Euler Mac Laurin Summation Formula, and this will give that \(\displaystyle\sum_{k=1}^{n}k^{-1/2}=2n^{1/2}+\zeta(1/2)+o(1)\) . This answers both questions. For details one can see Wilf's Mathematics for the Physical Sciences p.124 here.
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