# Convergence of infinite series-1

To investigate the convergence/divergence of this series:

$$\sum_{n=1}^{\infty} (n^{-1/2}-\sin(n^{-1/2}))^{1/2}$$

So, I took the maclaurin series$$\sin(x)\sim x-\frac{x^3}{3!}$$ Replacing it in the initial equation we get:

$$\sum_{n=1}^{\infty} (n^{-3/2})^{1/2}$$

which is divergent, by p-series since p=3/4<1.

Is this sufficient to say that the original series is divergent?

• No, you have to make pore precise the sign $\sim$. – Did Oct 9 '13 at 15:16
• I don't think I understand what you mean by "pore precise" – John Oct 9 '13 at 15:19
• He means "more precise", it was a typo. You must make sufficiently precise what $\sin x \sim x - \frac{x^3}{3!}$ means to draw the conclusion. – Daniel Fischer Oct 9 '13 at 15:28
• Something like "$\frac{x^3}{12} < x - \sin x < \frac{x^3}{6}$ for $0 < x < 1$", which would suffice. – Daniel Fischer Oct 9 '13 at 15:36
• Thanks for your help :)!! Just to be on the save side, I should include the landau symbol to keep track of how many terms I need and to be more precise. – John Oct 9 '13 at 16:26

From the power series expansion of $\sin x$, we conclude that for $0\lt x\le 1$, we have $$\sin x\lt x-\frac{x^3}{3!}+\frac{x^5}{5!}.$$ Thus $$\sin x \lt x-x^3\left(\frac{1}{6}-\frac{1}{120}\right).$$ It follows that $$n^{-1/2}-\sin(n^{-1/2})\gt \frac{19}{120}n^{-3/2},$$ and therefore if $a_n$ is the $n$-th term of our series, then $$a_n \gt \left(\frac{19}{120}\right)^{1/2}\frac{1}{n^{3/4}}.$$ By Comparison, our series diverges.
Remark: The analysis in the OP focused immediately on the right thing, the $x^3$ term in the power series of $\sin x$. In many contexts the analysis would be sufficient. The difference $n^{-1/2}-\sin(n^{-1/2})$ "behaves like" $n^{-3/2}$, so $a_n$ behaves like $n^{-3/4}$, which goes to $0$ too slowly for convergence.