Test convergence of the series $\sum \frac{k\sin^2 k}{1+k^3}$

Expert intro: This is a positive-term series with a trigonometric factor. The denominator grows like $k^3$, so the comparison is the main idea.

Detailed walkthrough

We study

$\sum_{k=1}^{\infty} \frac{k\sin^2 k}{1+k^3}.$

Step 1: Bound the trigonometric part

Since

$0\le \sin^2 k \le 1,$

we get

$0\le \frac{k\sin^2 k}{1+k^3} \le \frac{k}{1+k^3}.$

Step 2: Compare with a simpler series

For large $k$,

$\frac{k}{1+k^3} \sim \frac{k}{k^3}=\frac{1}{k^2}.$

So we compare with

$\sum \frac{1}{k^2},$

which converges.

Step 3: Apply comparison

Because

$0\le \frac{k\sin^2 k}{1+k^3} \le \frac{k}{1+k^3},$

and

$\sum_{k=1}^{\infty} \frac{k}{1+k^3}$

converges by comparison with $\sum 1/k^2$, the given series also converges.

Final answer

$\sum_{k=1}^{\infty} \frac{k\sin^2 k}{1+k^3} \quad \text{converges.}$

💡 Pitfall guide

A typical slip is to treat $\sin^2 k$ as if it averaged to something exact and then overcomplicate the problem. You do not need that here. The clean move is simply $\sin^2 k\le 1$. Also, don’t compare it to $1/k$; that is too weak and may mislead you.

🔄 Real-world variant

If the numerator were $k^m\sin^2 k$ instead, the same idea would compare the terms to $k^{m-3}$. Then the series would converge when $m-3<-1$, यानी $m<2$. So the exponent on $k$ in the numerator is what really controls the outcome.

🔍 Related terms

comparison test, bounded trigonometric function, p-series