I need to find whether this series is converging or diverging $$\sum_{n=1}^{\infty}\left(\frac{1}{n^{0.51}} - \sin(\frac{1}{n^{0.51}})\right)$$ I think that it is converging so I tried to prove the convergence:
I know that: $\sin(\frac{1}{n^{0.51}}) < \frac{1}{n^{0.51}}$
So: $\frac{1}{n^{0.51}} - \sin(\frac{1}{n^{0.51}}) > 0$
Now I know that I can use comparison test, but I can't find right series to compare to.
I think they choose $0.51$ since the problem is designed for the beginner of calculus. Therefore, the identity $x-\sin{x}={1 \over 6}x^3+O(x^5)$ is set as an unknown stuff.
A more basic way of proving this is considering the series $\sum_{n=1}^{+\infty}{\big(\tan\big({1 \over n^{0.51}}\big)-\sin\big({1 \over n^{0.51}}\big)\big)}$. Since $\tan\big({1 \over n^{0.51}}\big) > {1 \over n^{0.51}}$ , the original series converges if $\sum_{n=1}^{+\infty}{\big(\tan\big({1 \over n^{0.51}}\big)-\sin\big({1 \over n^{0.51}}\big)\big)}$ converges. With the inequality \begin{align} \tan\big({1 \over n^{0.51}}\big)-\sin\big({1 \over n^{0.51}}\big) & =\tan\big({1 \over n^{0.51}}\big)\big(1-\cos\big({1 \over n^{0.51}}\big)\big) \\ & < {\sin\big({1 \over n^{0.51}}\big) \over \cos\big({1 \over n^{0.51}}\big)}\big(1-\sqrt{1-{1 \over (n^{0.51})^2}}\big) \\ & < {1 \over \cos\big({1 \over 1^{0.51}}\big)} \big({1 \over n^{0.51}}\big)\big(1-\big(1-{1 \over n^{0.51}}\big)\big) \\ & = {1 \over \cos\big({1 \over 1^{0.51}}\big)}{1 \over n^{1.02}} \end{align} which works for any $n \in \mathbb{N}$, it is easy to show $\sum_{n=1}^{+\infty}{\big(\tan\big({1 \over n^{0.51}}\big)-\sin\big({1 \over n^{0.51}}\big)\big)}$ converges.
Consider the following function in the range $[0,1]$ $$f(x)=x-\frac16 x^3 - \sin x $$ $$f'(x)=1-\frac12 x^2 - \cos x $$ $$f''(x)=- x + \sin x $$ $$f'''(x)=- 1 + \cos x $$ $$f(0)=f'(0)=f''(0)=f'''(0)=0$$ Note that, in the given range, $0 \leqslant \sin x , \cos x \leqslant 1$.
So $f'''(x) \leqslant 0$, so $f''(x) $ is decreasing.
So $f''(x) \leqslant 0$, so $f'(x) $ is decreasing.
So $f'(x) \leqslant 0$, so $f(x) $ is decreasing.
$$f(x) \leqslant 0$$ $$x-\frac16 x^3 - \sin x \leqslant 0$$ $$x - \sin x \leqslant \frac16 x^3$$ $$\frac{1}{n^{0.51}} - \sin \frac{1}{n^{0.51}} \leqslant \frac{1}{6 n^{1.53}}$$
and so your series is convergent by the comparison test.
By this we have $$\lim_{n\to\infty}\frac{\frac{1}{n^{0.51}} - \sin(\frac{1}{n^{0.51}}) }{\frac{1}{n^{0.51x 3}}}=\frac16$$
Thus there is $n_0$ such that for all $n\geq n_0$ we have
$$\Big|\frac{\frac{1}{n^{0.51}} - \sin(\frac{1}{n^{0.51}}) }{\frac{1}{n^{0.51x 3}}}-\frac16\Big|\leq \frac{1}{12}$$ that is for all $n\geq n_0$, $$\frac{1}{12n^{1,53}}\leq \frac{1}{n^{0.51}} - \sin(\frac{1}{n^{0.51}}) \leq \frac{1}{4n^{1,53}}$$
Passing to the sum gives
$$\frac{1}{12} \sum_{n\geq n_0}\frac{1}{n^{1,53}}\leq \sum_{n\geq n_0} \frac{1}{n^{0.51}} - \sin(\frac{1}{n^{0.51}}) \leq \frac{1}{4} \sum_{n\geq n_0}\frac{1}{n^{1,53}}<\infty$$
this implies the convergence of your series.
