convergence

**qzno** · December 9th 2009, 04:02 PM

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

**ANDS!** · December 9th 2009, 04:13 PM

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).

**Prove It** · December 9th 2009, 04:18 PM

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.

**qzno** · December 9th 2009, 04:57 PM

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

**Prove It** · December 9th 2009, 05:06 PM

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

**Krizalid** · December 9th 2009, 06:03 PM

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

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