Examine if the series converge $\sum_{n=1}^\infty \sin^2(\frac{1}{n})$.
For which $a$$\in \mathbb{R^+}$ the series converge $$\sum_{n=1}^\infty\frac{\sqrt{n^4+1}}{n^a}?$$ Also for which $a$ the series converge absolute?
For the first i proved it with the integral test and i found $Si(2)-\sin^2(1)$. Can we find another test to prove its convergence?
For the second, $b_n$ is decreasing and positive. Also we need $\lim_{n\to∞} b_n=0$ for Leibniz criteria to work. So we get $a>2$
For the absolute convergence we take $|b_n|\leq \frac{n^2}{n^a}=\frac{1}{n^{a-2}}$ and so we get $a>3$.
Is this right or am i wrong?
For the first note that
$$\sin^2(\frac{1}{n})\sim \frac1{n^2}$$
For the second note that it is not an alternating series but it is strictly positive, a good method can be to observe that
$$\frac{\sqrt{n^4+1}}{n^a}\sim \frac1{n^{a-2}}$$
then use limit comparison test.