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$\sum_{n=1}^{\infty}n^a\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$

I tried to use the D'Alembert ratio test and it didn't quite work out. The series inside is telescoping but I don't know if that information will be useful. Can someone help?

marcopopo
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    Please include your question in the body of the question, instead of putting it only in the title. – José Carlos Santos Dec 13 '20 at 15:55
  • We know that $\sum n^a$ converges when $a<-1$ so the whole question comes down to the behavior of $\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$ as $n\to\infty$ Find $b,c$ such that $\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\sim cn^b$ as $n\to\infty$ and you're done. – saulspatz Dec 13 '20 at 16:12

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We have $$ \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}} = \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} + \sqrt{n})} \sim \frac{1}{2n^{3/2}}. $$ Hence $a_n = n^{\alpha} \Bigl( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \Bigr) \sim \frac{1}{2n^{3/2-a}}$. The sum $\sum_n a_n$ is convergent iff $\sum_n \frac{1}{n^{3/2-a}}$ is convergent. Using integral test for convergence we get that $\sum_n \frac{1}{n^{3/2-a}}$ is convergent iff $3/2-a > 1$. So, the sum $\sum_n a_n$ is convergent iff $a<\frac12$.

Botnakov N.
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