My professor asked me to prove that the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is convergent and find the sum.

But I ended up proving it divergent. Here's my work :

Notice that $\sqrt{1+\frac{1}{n}}<\frac{4}{\sqrt{2}}-1$ for all positive integer $n$. This implies $\sqrt{2n}\times\sqrt{n+1}<4n-\sqrt{2}\times n$ for all positive integer $n$, i.e., $\sqrt{2n}\times(\sqrt{n}+\sqrt{n+1})<4n$ for all positive integer $n$.

Finally, we get $\frac{1}{4n}<\frac{1}{\sqrt{2n}(\sqrt{n}+\sqrt{n+1})}=\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ for all positive integer $n$.

Since $\sum_{n=1}^{\infty}\frac{1}{4n}$ is divergent, it follows that $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is divergent by Comparison test.

Can someone tell me what's wrong with my proof ? Is the series really DIVERGENT ?

  • $\begingroup$ Nice proof. I think there is nothing wrong in what you wrote. $\endgroup$ – Ahmed Jun 2 '17 at 2:56
  • $\begingroup$ If you have polynomial expressions of radicals you can count the order of the terms, and then use limit comparison test. Messy proofs are not necessary $\endgroup$ – clark Jun 2 '17 at 4:16

We have $$2n\le 2n+2$$ and $$n\le 2n+2$$ thus

$$u_n\ge \frac {1}{\sqrt {2n+2}(\sqrt {2n+2}+\sqrt {2n+2})} $$ $$\ge \frac {1}{2 (2n+2)} $$

the series $\sum \frac {1}{2 (2n+2)} $ diverges and so does $\sum u_n $.


Yes the series diverges,

as $2n + 2 \geq n $ and $\sqrt{2n} \leq 2\sqrt{n}$,so $\sqrt{2n} + \sqrt{2n+2} \leq 2\sqrt{n} + \sqrt{n} = 3 \sqrt{n}$,

So,$\frac{1}{(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3\sqrt{n}}$,

$\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3n}$ or $\frac{1}{3n} \leq \frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ ,so by the comparision test the series diverges.


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