# Homework Help: Series containing trigonometry

1. Mar 10, 2014

### dmcmu

1. The problem statement, all variables and given/known data
Determine whether the series [3+cos(n)]/n from n=1 to infinity is convergent or divergent. Explain your reasoning and demonstrate that any conditions required to use a particular test are satisfied.

2. Relevant equations
Any series tests: ratio test, alternating sequence test

3. The attempt at a solution
After writing out the first few terms I believe that the series is alternating and converging, however I can't seem to find out how to convert this into the form of an alternating series to be able to perform the alternating series test. I know there is a way to use power series to convert cosine to a series form but I believe this question is meant to be done without power series. Thanks in advance.

2. Mar 10, 2014

### jbunniii

$3 + \cos(n)$ is positive for all $n$, so it can't be an alternating series.

Try finding some simple upper and lower bounds for $3 + \cos(n)$, and see if you can conclude anything.

3. Mar 10, 2014

### SteamKing

Staff Emeritus
And you can also split the given series into two separate series and see if both are convergent separately.

4. Mar 10, 2014

### Dick

I wouldn't do that. cos(n)/n converges, but the proof of that is considerably harder than proving the original series diverges.

5. Mar 10, 2014

### SteamKing

Staff Emeritus
But does the series (3/n) converge? After all, I was only suggesting taking (3 + cos (n))/n and re-writing it as (3/n) + (cos (n)/n). Since, as was stated above, all terms of this series are positive and hence non-alternating, you should be able to examine the convergence of the two resulting series independently. If one series can be shown to diverge, it doesn't really matter what the other series does.

6. Mar 10, 2014

### Curious3141

But $\cos n$ is negative infinitely often. If indeed that summation diverges, it may be to negative infinity. In which case, you're still left with the difficult task of deciding how that "offsets" the positively divergent harmonic sum part.

The only way to conclude that the entire series is divergent after splitting them up is to recognise that the series sum of $\frac{\cos n}{n}$ is convergent. In which case, the divergence of the harmonic sum part makes for a trivial conclusion. But showing that trig part is convergent is the real toughie.

Much better to just bound the entire terms of the series by constant multiples of harmonic terms and use the squeeze theorem.

7. Mar 10, 2014

### Dick

The sum of two divergent series can converge. (-1)^n diverges, (-1)^(n+1) diverges. The sum is zero and converges. As Curious3141 said, if you can show cos(n)/n converges then the divergence of the harmonic part is enough, but that requires something like a Dirichlet test or a partial sum using Euler's equation. Way harder than showing the original series diverges.

Last edited: Mar 10, 2014
8. Mar 11, 2014

### SteamKing

Staff Emeritus
I see what you're saying now. My thinking cap slipped off my head.

9. Mar 11, 2014

### Curious3141

Has been increasingly happening to me of late. Must be the thinning hair. :rofl:

10. Mar 11, 2014

### dmcmu

Oh! I see now, thanks for the help.