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I am trying to test the convergence of the series

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

$p$ is a fixed real number.

I tried the ratio test and the integral test without success. Now, I am stuck with the quantity between the round brackets. I have no idea on how to treat it in order to obtain something useful to work with.

Did
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Charlie
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  • Is $p$ a prime? – mike Sep 02 '14 at 14:48
  • positive or negative? – mike Sep 02 '14 at 14:50
  • @mike Mr. Apostol says nothing about that... $p$ is just a fixed real number. – Charlie Sep 02 '14 at 14:52
  • In several recent questions you explain that you tried the ratio test and that it failed. Please try to recognize once and for all that when the general term of a series is equivalent to $c/n^a$ then the ratio test always fail. Consequence: it is time to learn other tools, such as the ones explained in the answers you already received. – Did Sep 02 '14 at 14:53
  • @Did God Bless you. I totally agree with you. Learning new ways is what I am looking for. I am working hard to improve my math skills. Thank you. – Charlie Sep 02 '14 at 14:59

2 Answers2

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Tool: For every fixed $a$, when $x\to0$, $(1+x)^a=1+ax+\frac12a(a-1)x^2+o(x^2)$. In particular, the case $a=\frac12$ yields $\sqrt{1+x}=1+\frac12x-\frac18x^2+o(x^2)$.

For every fixed $c$, $$\sqrt{n+c}=\sqrt{n}\cdot\sqrt{1+\frac{c}n}=\sqrt{n}\cdot\left(1+\frac{c}{2n}-\frac{c^2}{8n^2}+o\left(\frac1{n^2}\right)\right),$$ hence $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}=\sqrt{n}\cdot\left(1+\frac1{2n}-\frac1{8n^2}-2+1-\frac1{2n}-\frac1{8n^2}+o\left(\frac1{n^2}\right)\right),$$ that is, $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\sim\sqrt{n}\cdot\frac{-1}{4n^2}=\frac{-1}{4n^{3/2}}.$$ Can you finish this?

Did
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In essence the problem is to determine the order of $$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}$$

Tring the ususal trick of multiplying by the conjuguate we get

$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n} =\frac{2\sqrt{n^2-1}-2n}{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}$$ then doing the same again we get

$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n} =\frac{-2}{(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n})(\sqrt{n^2-1}+n)}$$

Now the order of $(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n})(\sqrt{n^2-1}+n)$ is $n^{\frac{3}{2}}$ as can be easily verified, since $$\frac{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}{\sqrt{n}}\rightarrow 4$$ and

$$\frac{\sqrt{n^2-1}+n}{n}\rightarrow 2$$

So if we compare our series with the series $\sum n^{p-\frac{3}{2}}$ we see that we get a finite ratio. Thus the criteria for convergence is

$$p-\frac{3}{2}<-1$$

or $$p<\frac{1}{2}$$

Rene Schipperus
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  • I think there is something wrong in your conclusion. The criteria for convergence should be $\frac{3}{2}-p>1 ⇒ p<1/2$. Anyway I accepted your answer because I need more time to completely understand what @Did wrote. Thank you. – Charlie Sep 02 '14 at 21:23
  • Sorry its just a sign error, I edited. – Rene Schipperus Sep 02 '14 at 23:14