Does series converge absolutely or conditionally?

Question

I was wondering if you guys could judge my reasoning and let me know if am correct in finding if this series converges absolutely or conditionally.

$$\sum^{∞}_{k=1} \frac{\sin(2k^2+1)}{k^{3/2}}$$

I took that absolute value and since $\sin(2k^2+1)$ is bounded, the summand should “look like” $\frac{const}{k^{3/2}}$:

$$\sum^{∞}_{k=1} \frac{|\sin(2k^2+1)|}{k^{3/2}} \le \sum^{\infty}_{k=1} \frac{1}{k^{3/2}}$$

Since the summand is positive, we can use the Comparison Test, since |\sin(2k^2+1)|<1 for all k, we have:

$$\sum^{\infty}_{k=1} \frac{|\sin(2k^2+1)|}{k^{3/2}} < \sum^{∞}_{k=1} \frac{1}{k^{3/2}}$$

and since $\sum^{∞}_{k=1} \frac{1}{k^{3/2}}$ converges by p-series, $\sum^{\infty}_{k=1} \frac{|\sin(2k^2+1)|}{k^{3/2}}$ converges. So the original series $\sum^{\infty}_{k=1} \frac{\sin(2k^2+1)}{k^{3/2}}$ converges absolutely

Answer 1

Yes, your reasoning is correct, and the series converges absolutely.

The reasoning can be written concisely as follows.

Since for each positive integer $k$, $$ \left|\frac{\sin(2k^2+1)}{k^{3/2}}\right|\le \frac{1}{k^{3/2}} $$ and the $p$-series $\displaystyle\sum \frac{1}{k^{3/2}}$ converges, by the M-test, $\displaystyle \sum\left|\frac{\sin(2k^2+1)}{k^{3/2}}\right|$ is convergent. Thus the original sereies converges absolutely.