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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?
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.
Once one knows that the intuitive argument can be fleshed out to give a full proof, there is no need to give the details. However, at this stage, part of the point is to acquire enough experience so that later one can legitimately leave out details.
