convergence/divergence of (1+cosn)/n^2

[petrol.veem]

I've been working in the integral test and comparison test section recently and came across this problem:

Sum(1,infinity) [ (1 + cos(n)) / n^2 ]

So the first thing that i noticed is that the top is bounded by (0,2). I wanted to try to compare it to something like 1 / n^2 but this seems like it wouldn't work since the one I'm looking at can go above 1.

Then I tried the integral test, but integrating a function of that nature also isn't very easy I found.

So now I'm not exactly sure where to go...

 

[pka]

Sum(1,infinity) [ (1 + cos(n)) / n^2 ]
So the first thing that i noticed is that the top is bounded by (0,2). I wanted to try to compare it to something like 1 / n^2 but this seems like it wouldn't work since the one I'm looking at can go above 1.

\(\displaystyle \[\frac{{1 + \cos (n)}}{{n^2 }} \leqslant \frac{2}{{n^2 }}\)
Of course it works. It converges by direct comparison.

 

[pka]

Sum(1,infinity) [ (1 + cos(n)) / n^2 ]
So the first thing that i noticed is that the top is bounded by (0,2). I wanted to try to compare it to something like 1 / n^2 but this seems like it wouldn't work since the one I'm looking at can go above 1.

\(\displaystyle 0 \le \frac{{1 + \cos (n)}}{{n^2 }} \leqslant \frac{2}{{n^2 }}\)
Of course it works. It converges by direct comparison.

 

[PAULK]

I've been working in the integral test and comparison test section recently and came across this problem:

Sum(1,infinity) [ (1 + cos(n)) / n^2 ]

So the first thing that I noticed is that the top is bounded by (0,2).

>> That's it! That means:
inf 1 + cos n
SUM ------------- <=
n=1 n^2

inf | 1 + cos n |
SUM ----------------- <=
n=1 n^2

inf 2
SUM -------------- =
n=1 n^2

inf 1
2 SUM --------- =
n=1 n^2

which converges.

I wanted to try to compare it to something like 1 / n^2 but this seems like it wouldn't work since the one I'm looking at can go above 1.

As you see, that does not matter. As long as it is bounded, it's dominated by the upper bound.