# converges or diverges?

I have the following series:

$\sum_n{a_n}$, where $a_1 = 1$, $a_{n+1} = a_n \,\frac{2 + \cos n}{\sqrt n}$. Does it converge or diverge?

Edit: formatted.

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Please do use LaTeX to properly write mathematics. I tried to edit your post but it confuses me. In FAQ you can find some directions for this. –  DonAntonio Nov 20 '12 at 5:17
@DonAntonio You got it right! I did not know it was using LaTeX at all. May you help me to figure it out now?> –  philippe Nov 20 '12 at 5:19

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I already the Ratio test. I now need to know if cos(n)/sqrt(n) diverges or converges –  philippe Nov 20 '12 at 5:26
As a sequence, it converges to $0$. The numerator is bounded and the denominator is unbounded and increasing. –  alex.jordan Nov 20 '12 at 5:28
I applied the Ratio test: I got (2 + cos( n + 1 )) / (n + 1 ). As this limit goes to 0 when n-> infinite, then the series a_n also converges? –  philippe Nov 20 '12 at 5:30
Yep. If the limit of the ratio of a term to the previous term is less than 1, then the terms can be summed to a finite sum. –  alex.jordan Nov 20 '12 at 5:33

Consider the sequence $a_{n+1}=a_n\frac{3}{\sqrt{n}}$. See whether it converges or not. Now see if this converges, why should the original sequence converge.

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I got 3 / sqrt( n + 1 ) when I applied the ratio test for the sequence you gave me. Am I right? –  philippe Nov 20 '12 at 5:34
You have to see the limit for the ratio test. What is the ratio $|\frac{a_{n+1}}{a_n}|$ at the limit $n$ tends to infinity. Now see that each term of the original sequence is upper bounded by each term in this sequence. –  dineshdileep Nov 20 '12 at 5:38

Continuing with Alex's answer: apply now Dirichlet's test*, with the monotone descending sequence $\,\displaystyle{\left\{\frac{1}{\sqrt n}\right\}}\,$ and the bounded sequence $\,\displaystyle{\left\{\sum_{k=1}^n\cos n\right\}}\,$

• cazelais.disted.camosun.bc.ca/250/dirichlet.pdf
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Alternatively, use direct comparison with a geometric series, noting that all terms are positive. The first 15 terms sum to whatever they sum to. After that, with $n\ge16$, then $a_{n+1}=a_n\frac{2+\cos(n)}{\sqrt{n}}\le\frac{3}{4}a_n$. Inductively, $a_n\le\left(\frac34\right)^{n-16}a_{16}$ for $n\geq16$. So the series sums to at most $$\sum_{n=1}^{15}a_n+a_{16}\sum_{n=16}^\infty\left(\frac34\right)^{n-16}$$ or rather $$\sum_{n=1}^{15}a_n+4a_{16}$$

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