# Math Help - convergence

1. ## convergence

Use an appropriate test to determine the absolutely convergence, conditionally convergence or divergence of the series. Identify the test used.

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

This was on a past test and my teacher doesn't provide us with the answers so can someone help solve this for me

2. This problem is easier than you think. The series doesn't alternate (it will always be positive - you can figure out why on your own). The function f(n) also aproaches zero, as n tends to infinity as the numerator will always be a number between 1 and 0 (there maybe some fluctuation - but you are approaching zero). You can use the comparison test, and notice that the similar function is 1/2^n (which is convergent).

3. Originally Posted by qzno
Use an appropriate test to determine the absolutely convergence, conditionally convergence or divergence of the series. Identify the test used.

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

This was on a past test and my teacher doesn't provide us with the answers so can someone help solve this for me
It's fairly long, but first, it's a positive termed sequence, so you don't need to test for absolute or conditional convergence. Just test for convergence.

Also note that $0 \leq \cos^2(n + 1) \leq 1$.

This would mean that

$\frac{\cos^2(n + 1)}{2^n + 1} \leq \frac{1}{2^n + 1}$.

We can say that if the "larger" series converges, so must the "smaller" series. So use the comparison test.

$\sum_{n = 2}^\infty\frac{\cos^2(n + 1)}{2^n + 1} \leq \sum_{n = 2}^\infty\frac{1}{2^n + 1}$.

Now applying the ratio test...

$a_k = \frac{1}{2^k + 1}$

$a_{k + 1} = \frac{1}{2^{k + 1} + 1}$

$\frac{a_{k + 1}}{a_k} = \frac{2^k + 1}{2^{k + 1} + 1}$.

$= \frac{1}{2} + \frac{1}{2\left(2^{k + 1} + 1\right)}$.

Now letting $k \to \infty$ we see

$\lim_{k \to \infty}\frac{a_{k + 1}}{a_k} = \frac{1}{2}$.

Since this limit $<1$, the series must converge.

And since our "larger" series converges, so must the "smaller" series.

4. Thank you for that but I'm just wondering if there's any other way to figure it out of do you have to use those specific methods

5. Since you've already been shown two correct methods, I'd say there are lots of methods that can be used.

6. note that $\frac{1}{2^{n}+1}<\frac{1}{2^{n}},$ so direct comparison applies faster.