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 $$0.51 > 0.5$$.

• Hi thanks for the response! We haven't gone through Taylor Series yet unfortunately. :( May 10, 2021 at 13:11
• The series converges iff $p\gt\frac13$. What you've shown is that the series converges if $p\gt\frac12$.
– robjohn
May 10, 2021 at 22:20

You can use the fact that for $$x \geq 0$$, $$x - \sin(x) \leq \frac{1}{6} x^3.$$

• How would one prove such a claim? May 10, 2021 at 13:12
• @MathCurious Show that $f(x) = \frac16 x^3 - x + \sin x \ge 0$ May 10, 2021 at 13:21
• So how do you prove that the RHS function converges? May 10, 2021 at 14:05
• Yes, that is what I'm asking May 10, 2021 at 14:28
• @MathCurious: this answer gives a pre-calculus proof that $\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$.
– robjohn
May 10, 2021 at 21:04

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 $$\displaystyle{\sum_{n=1}^{\infty}b_{n}}}$$.