\(\sum_{n=1}^{\infty} n^a(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})\)

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Grigorios Kostakos
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\(\sum_{n=1}^{\infty} n^a(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})\)

#1

Post by Grigorios Kostakos »

Examine the convergence of the series \[\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}{n^{a}\bigl({\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\,}\bigr)}\] for all real numbers \(a\) .
Last edited by Grigorios Kostakos on Wed Nov 11, 2015 4:16 am, edited 2 times in total.
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Re: \(\sum n^a(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})\)

#2

Post by akotronis »

We have

\(\begin{aligned}n^a(\sqrt{n+1}-2\sqrt{2}+\sqrt{n+1})&=n^{a+1/2}(\sqrt{1+1/n}-2+\sqrt{1-1/n})\\&=n^{a+1/2}(1+\frac{1}{2n}-\frac{1}{8n^2}-2+1-\frac{1}{2n}-\frac{1}{8n^2}+\mathcal O(n^{-3}))\\&=-\frac{1}{4n^{3/2-a}}+\mathcal O(n^{a-5/2})\,,\end{aligned}\)
so the series converges for \(a<1/2\) and is \(-\infty\) for \(a\geq1/2\).
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Re: \(\sum n^a(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})\)

#3

Post by Grigorios Kostakos »

... and an elementary way:
\begin{align*}
& \displaystyle\mathop{\sum}\limits_{n=1}^{\infty}{n^{a}\bigl({\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\,}\bigr)}=\\
& \displaystyle\sqrt{2}-2+\mathop{\sum}\limits_{n=2}^{\infty}{n^{a}\biggl({\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n}+\sqrt{n-1}}}\biggr)}=\\
& \displaystyle\sqrt{2}-2+\mathop{\sum}\limits_{n=2}^{\infty}{n^{a}\,\frac{\sqrt{n-1}-\sqrt{n+1}}{\bigl({\sqrt{n+1}+\sqrt{n}\,}\bigr)\bigl({\sqrt{n}+\sqrt{n-1}\,}\bigr)}}=\\
& \displaystyle\sqrt{2}-2-2\,\mathop{\sum}\limits_{n=2}^{\infty}{\frac{n^{a}}{\bigl({\sqrt{n+1}+\sqrt{n}\,}\bigr)\bigl({\sqrt{n}+\sqrt{n-1}\,}\bigr)\bigl({\sqrt{n+1}+\sqrt{n-1}\,}\bigr)}}
\end{align*}
\(\bullet\) If \(a<\frac{1}{2}\), then \begin{align*}
& 0>\displaystyle\sqrt{2}-2-2\mathop{\sum}\limits_{n=2}^{\infty}{\frac{n^{a}}{\bigl({\sqrt{n+1}+\sqrt{n}\,}\bigr)\bigl({\sqrt{n}+\sqrt{n-1}\,}\bigr)\bigl({\sqrt{n+1}+\sqrt{n-1}\,}\bigr)}}\\
& >-\mathop{\sum}\limits_{n=1}^{\infty}{\frac{n^{a}}{n^{\frac{3}{2}}}}=-\mathop{\sum}\limits_{n=1}^{\infty}{\frac{1}{n^{\frac{3}{2}-a}}}=-\zeta\bigl({\tfrac{3}{2}-a}\bigr)\,.
\end{align*}
\(\bullet\) If \(a\geqslant\frac{1}{2}\), then \begin{align*}
&\displaystyle\sqrt{2}-2-2\mathop{\sum}\limits_{n=2}^{\infty}{\frac{n^{a}}{\bigl({\sqrt{n+1}+\sqrt{n}\,}\bigr)\bigl({\sqrt{n}+\sqrt{n-1}\,}\bigr)\bigl({\sqrt{n+1}+\sqrt{n-1}\,}\bigr)}}\\
&<\displaystyle\sqrt{2}-2-2\mathop{\sum}\limits_{n=2}^{\infty}{\frac{n^{a}}{8\,({n+1})^{\frac{3}{2}}}}=\sqrt{2}-2-\frac{1}{4}\,(+\infty)=-\infty\,.
\end{align*}
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Re: \(\sum n^a(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})\)

#4

Post by Demetres »

We will use the following generalisation of the mean value theorem (see e.g. Problem V98 in "Problems and Theorems in Analysis II" by Pólya and Szegő):

If \(f\) is \(n\)-times differentiable and \(h > 0\) then \[ \frac{1}{h^n}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}f(x+kh) = f^{(n)}(\xi)\] for some \(\xi \in (x,x+nh)\)

In particular, taking \(f(x) = \sqrt{x},h=1\) and \(n=2\) we have \[ \sqrt{n+1} - 2\sqrt{n} + \sqrt{n-1} = -\frac{\xi_n^{-3/2}}{4} \] for some \(\xi_n \in (n-1,n+1)\).

An easy comparison test with series of the form \(\sum n^{\beta}\) (whose convergence properties are well-known) shows that the original series converges if and only if \(a < 1/2\)
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