Question

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Answer 1

\[\sum_{n=1}^\infty \frac{2+\cos n}{n^2}\]
Use the fact that cosine is bounded.
\[-1\le\cos n\le1~~\Rightarrow~~1\le\cos n+2\le3\]
So, whatever \(n\) may be, the numerator of the series will be a constant between \(1\) and \(3\).

It doesn't matter what number the numerator takes on, so call it \(c\). Because of this you can compare it to the series \(\sum\dfrac{1}{n^2}\), which converges.

To apply the test, you must show that \(\dfrac{1}{n^2}\le\dfrac{c}{n^2}\), which should be quite obvious, since \(1\le c\le3\).

Answer 2

How do I show that c/(n^2) is greater or equal to it? Just by plugging in the numbers from 1-3?

Answer 3

If \(c=1\), then of course \(\dfrac{1}{n^2}\le\dfrac{1}{n^2}\). Any value of \(c>1\) will also satisfy this. All you have to consider is the smallest value \(c\) can take on.

Answer 4

So since anything from 1-3 works, then the series converges?

Answer 5

Yes

Answer 6

Oh ok, perfect, thanks! ^^

Answer 7

yw