Prove that the divergence of the following sequence. $$s_n= -2n^{\frac{9}{8}}$$
To prove this I was thinking of starting with $$n^{\frac{9}{8}} > n$$
Then, $$2n^{\frac{9}{8}} > 2n$$
Then, $$-2n^{\frac{9}{8}} < -2n$$.
Then proving that $$\lim_{n\to\infty}-2n = -\infty$$
And since $$-2n^{\frac{9}{8}} < -2n$$, by comparison, $$s_n= -2n^{\frac{9}{8}}$$ diverges.
I know there is more than one way to prove divergent of a sequence such as using contradiction of convergence using epsilon... Etc. But does my method count as a valid proof? I also tried proving this using the odd and even integer subsequence to prove this but I got stuck. Thanks.
It is completely correct. In the comments, I provided a proof for the fact:
$$a_n < b_n, b_n \to -\infty \implies a_n \to -\infty$$ which you used implicitely in your proof.