I'm trying to prove this series converges by using some sort of comparison test.
$$\sum_{n=1}^{\infty}\frac{1}{n^{0.51}}-\sin\left(\frac{1}{n^{0.51}}\right)$$
I know by $\sin n\le n$ that the series is positive, so I went with the direction of using the comparison test. But I can't seem to find a function that is always greater than the expression in the series that also converges..
Using the Taylor series for $\sin x$, we have $$\sin \frac{1}{n^p} = \frac{1}{n^p} + O\left(\frac{1}{n^{2p}}\right)$$ Therefore $$S=\sum_{n\ge 1}\frac{1}{n^p} - \sin \frac{1}{n^p} = \sum_{n\ge 1}\frac{1}{n^p} - \frac{1}{n^p} + O\left(\frac{1}{n^{2p}}\right) = \sum_{n\ge 1}O\left(\frac{1}{n^{2p}}\right)$$ Hence, the series $S$ converges if $p > 0.5$.
Your series converges since $0.51 > 0.5$.
You can use the fact that for $x \geq 0$, $$x - \sin(x) \leq \frac{1}{6} x^3.$$
Using Heine's definition for limits at $\infty$ and using the limit comparison test with $\;b_n=\frac{1}{6\left(n^{0.51}\right)^{3}}=\frac{1}{6n^{1.53}}$ , we get:
$$\lim_{n\to\infty}\frac{\frac{1}{n^{0.51}}-\sin\left(\frac{1}{n^{0.51}}\right)}{\frac{1}{6n^{1.53}}}{=}\lim_{x\to\infty}\frac{\frac{1}{x^{0.51}}-\sin\left(\frac{1}{x^{0.51}}\right)}{\frac{1}{6x^{1.53}}}\underset{t=\frac{1}{x^{0.51}}}{=}\lim_{t\to0^{+}}\frac{t-\sin t}{\frac{t^{3}}{6}}\underset{L}{=}$$ $$\lim_{t\to0^{+}}\frac{1-\cos t}{\frac{t^{2}}{2}}\underset{L}{=}\lim_{t\to0^{+}}\frac{\sin t}{t}=1$$ Hence $\displaystyle{\sum_{n=1}^{\infty}\left[\frac{1}{n^{0.51}}-\sin\left(\frac{1}{n^{0.51}}\right)\right]}$ converges with $\displaystyle{{\displaystyle \sum_{n=1}^{\infty}b_{n}}}$.