Find the value of $a$ such that the series converges.

I have to find the value of a such that the following series converges:$$\sum_{n=1}^{\infty}n^{\frac{1}{3}}\left|\sin\left(\frac{1}{n}\right)-\frac{1}{n^a}\right|$$

First of all, I simplified the series with the respective asymptotic as follows: $$\sum_{n=1}^{\infty}n^{\frac{1}{3}}\left|\frac{1}{n}-\frac{1}{n^a}\right| \sim \sum_{n=1}^{\infty}n^{\frac{1}{3}}\left|\frac{1}{n^a}\right|$$ Then, simplifying again, it results: $$\sum_{n=1}^{\infty}\frac{1}{n^{a-\frac{1}{3}}}$$ Which should converge for $$a>\frac{4}{3}$$, but I checked with Mathematica, which according to it, it is wrong.

• Uhm... Are you sure $\left\lvert \frac1n-\frac1{n^\alpha}\right\rvert\sim \frac1{n^\alpha}$? That ain't looking right. – Saucy O'Path Apr 13 at 15:38
• As @SaucyO'Path mentions, $\frac{1}{n} > \frac{1}{n^a}$ for all $a > 1$. Therefore, you could make a distinction into two cases, $a > 1$ and $a > 1$ and then you could get rid of the absolute value. – Viktor Glombik Apr 13 at 15:39
• Mathematica is like a gun. Sure, it comes in handy every now and then, but it ain't no smarter than the dude using it. – Ivan Neretin Apr 13 at 15:40
• Also, a very special thing happens at $a=1$. – Ivan Neretin Apr 13 at 15:42

(First of all, $$a_n\sim b_n$$ and $$\sum_{n=1}^\infty a_n \sim \sum_{n=1}^\infty b_n$$ are not the same thing; the second, actually, does not mean anything. But that's not the main point...)
Note that, as $$n\to\infty$$, $$\sin\frac{1}{n} - \frac{1}{n^a} = \frac{1}{n} - \frac{1}{6n^3} - \frac{1}{n^a} + o\left(\frac{1}{n^3}\right)$$ so that $$n^{1/3}\left(\sin\frac{1}{n} - \frac{1}{n^a} \right) = \left(\frac{1}{n^{2/3}} - \frac{1}{n^{a-1/3}} \right) - \frac{1}{6n^{8/3}} + o\left(\frac{1}{n^{8/3}}\right)$$ So for the series $$\sum_n n^{1/3}\left|\sin\frac{1}{n} - \frac{1}{n^a} \right|$$ to converge, you need the term in parentheses to cancel (why?), which gives you what $$a$$ must be.
If you take $$a=1$$, then $$\left\lvert\sin\left(\frac1n\right)-\frac1n\right\rvert$$ behaves as $$\frac1{n^3}$$, since$$\lim_{n\to\infty}\frac{\left\lvert\sin\left(\frac1n\right)-\frac1n\right\rvert}{\frac1{n^3}}=\lim_{x\to0^+}\frac{\left\lvert\sin(x)-x\right\rvert}{x^3}=\frac16.$$So, take $$a=1$$ and use the fact that the series$$\sum_{n=1}^\infty\frac{\sqrt[3]n}{n^3}$$converges.