Does \(\sum \frac{\sin(2n)}{1+2^n}\) Converge?

Key concept: This one is easier than it looks. The sine term stays bounded between $-1$ and $1$, while the denominator grows exponentially.

Step by step

Step 1: Bound the numerator

Since

$-1\le \sin(2n)\le 1,$

we get

$\left|\frac{\sin(2n)}{1+2^n}\right|\le \frac{1}{1+2^n}.$

Step 2: Compare with a geometric series

Also,

$\frac{1}{1+2^n}\le \frac{1}{2^n}.$

So

$\left|\frac{\sin(2n)}{1+2^n}\right|\le \frac{1}{2^n}.$

The geometric series

$\sum_{n=1}^{\infty}\frac{1}{2^n}$

converges.

Step 3: Conclude absolute convergence

By comparison,

$\sum_{n=1}^{\infty}\frac{\sin(2n)}{1+2^n}$

converges absolutely, hence it converges.

Why this works so quickly

The sine factor can oscillate, but it never gets large. The exponential denominator is what drives the terms to zero fast enough.

Pitfall alert

Don't try to use the alternating series test here. The terms are not of the simple form $(-1)^n b_n$, and you do not need that test anyway. Bounding $\sin(2n)$ and comparing to $1/2^n$ is cleaner.

Try different conditions

If the denominator were $1+n$ instead of $1+2^n$, the comparison would be much weaker and the series might fail to converge absolutely. If the numerator were $\cos(2n)$, the same boundedness argument would still work with no real change.

Further reading

absolute convergence, bounded sequence, geometric series