# Check for the convergence or divergence of the following series

1. Oct 4, 2013

### freshman2013

1. The problem statement, all variables and given/known data
Here are some series I'm completely stuck on.
1.sqrt(n)*(1-cos(1/n))

2. a series in which if n is odd, then an is 1/(n+$\sqrt[]{n}$) while if n is even, then an is -1/n

2. Relevant equations

3. The attempt at a solution
For 1., I tried integral test which seemed impossible to integrate, and then I tried comparison test but i can't find anything to compare to

For 2. I thought that as n approaches infinity for 1/(n+$\sqrt[]{n}$), which equals (1/$\sqrt{n}$)/($\sqrt{n}$+1), then that expression pretty much looks like 1/n and thus the series is approx (-1)^n/n and is thus conditionally convergent like (-1)^n/n. But I then realized perhaps you can't apply the alternating series test since in the original series, the abs value of it is not decreasing for every term.E.g.1/(100+root(100))=1/110 < 1/101 Anyone have any idea on how to tackle this, or am I right from my original attempt?

Last edited: Oct 4, 2013
2. Oct 4, 2013

### Zondrina

Your series is: $\sum_{n=1}^{∞} \sqrt{n} - \sqrt{n}cos(\frac{1}{n})$

I know this may seem misleading, but this series appears to converge.

Hint: $|cos(\frac{1}{n})| ≤ 1$

3. Oct 4, 2013

### haruspex

Yes, it converges, but you need to use a stronger fact about cos of small angles. Freshman2013, do you know the Taylor expansion of cos?

4. Oct 4, 2013

### jbunniii

For the first problem, this trig identity may be useful: $2\sin^2(x) = 1 - \cos(2x)$.

5. Oct 4, 2013

### jbunniii

For problem 2, try adding $a_n$ and $a_{n+1}$, where $n$ is odd.

6. Oct 4, 2013

### freshman2013

so for the second one, could I do this:
the first term 1/(n+$\sqrt{n}$) for all odd n, I rewrite as a separate series 1/(2n-1+$\sqrt{2n-1}$) and the 2nd term -1/n for all even n I rewrite as -1/(2n)

I combine those two series and I get (1-$\sqrt{2n-1}$)/(4n^2-2n+2n$\sqrt{2n-1}$) and test for that(which is convergent). And I do the same except with 1/2n instead -1/2n to check for absolute convergence.

7. Oct 4, 2013

### jbunniii

For absolute convergence, you can do something simpler. I claim that if there is some subsequence $(n_k)_{k=1}^{\infty}$ such that $\sum_{k=1}^{\infty} |a_{n_k}|$ diverges, then $\sum_{n=1}^{\infty} |a_n|$ must also diverge. If you can prove this, then you can apply it immediately to your series.