Alternating Series - Error Bound: $\sum_{n=1}^\infty\frac{2(-1)^n}{n^{0.9}}$

Question

Evaluating the series $$ \sum_{n=1}^\infty\frac{2(-1)^n}{n^{0.9}} $$ with an $|\text{error}|<0.001$.

I know that you would have to take the abs value which would then leave the problem at 2/(n^0.9).. That's all I know..

Answer 1

Hint. One may recall the alternating series test saying that if $|a_n|$ decreases monotonically and $\lim_{n \to \infty} a_n = 0$ then the alternating series converges. Moreover, with $$ S_N = \sum_{n=1}^N (-1)^{n-1} |a_n|, \quad S_\infty = \sum_{n=1}^\infty (-1)^{n-1} |a_n| $$ one has $$ \left|S_\infty-S_N\right|\le \left| a_{N+1}\right| $$ then one may apply it to $a_n=\dfrac2{n^{0.9}}$, giving $$ \dfrac2{n^{0.9}}<0.001 \implies n> 2000^{1/0.9}\approx \color{red}{4654}. $$