Determine whether $\sum _n \frac{{1+\sin (n^2) e^{\cos n}}}{n^\frac{3}{2}}$ is convergent or divergent

Question

Determine whether the series convergent or divergent. $$\sum _n \frac{{1+\sin (n^2) e^{\cos n}}}{n^\frac{3}{2}}$$

How do we tell whether the series is convergent or divergent, and what test we are going to use (root test, ratio, etc.)

Or, we can compare to $$\left|\frac{{1+\sin (n^2) e^{\cos n}}}{n^\frac{3}{2}}\right|\le \frac{1}{n^\frac{3}{2}}$$ So convergent?

Answer 1

$1\over {n^{3\over 2}}$ converges, $|\sin(n^2)e^{\cos(n)}|\leq e$, we deduce that $|\sin(n^2)e^{\cos(n)}|\over {n^{3\over 2}}$ $\leq$ $e\over {n^{3\over 2}}$ and the series converges.

Answer 2

It should be convergent.

Let's consider sequence $a_n=\sum_{k=1}^{n}\cfrac{3+\sin(k^2)e^{\cos k}}{k^{\frac{3}{2}}}$.

we know $\lim_{n\to\infty} a_n$ has a upper bound as it's less than $\sum_{n=1}^{\infty}\cfrac{3+e}{n^{\frac{3}{2}}}$;

we also know $\{a_n\}$ is increasing as each added term are positive.

$\therefore \{a_n\}$ is convergent.

$\sum_{n=1}^{\infty}\cfrac{1+\sin(n^2)e^{\cos n}}{n^{\frac{3}{2}}}=\lim_{n\to\infty}a_n-\sum_{n=1}^{\infty}\cfrac{2}{n^{\frac{3}{2}}}$, therefore it's convergent.