# How would you determine the convergence or divergence of this Series

1. Jul 21, 2009

### Bachelier

1. The problem statement, all variables and given/known data

sum 1/n*sin (1/n), n=1..infinity

I tried the limit comparison test, but I always get 0.
The ratio test is impossible
Comparison to the harmonic series cannot be used because 1/n*sin (1/n) is smaller than 1/n

Can you guys help?

Thanks

2. Jul 21, 2009

### zcd

Integral test yields a finite answer, so the series converges. For an easier solution you could compare the series to $$\frac{1}{n^{2}}$$
$$\lim_{n\to\infty}n\sin(\frac{1}{n})=1$$, and the geometric series converges.

Last edited: Jul 21, 2009
3. Jul 21, 2009

### Bachelier

I see where I made a mistake. I compared the series to 1/n^2 before but when calculating the limit of cos n at 0 (after applying l"hopital's rule), I put 0 as a result instead of 1.

Therefore the series converges since 1/n^2 converges.

Thanks for the help. :)

4. Jul 25, 2009

### Bachelier

Can you please explain the integral test especially in the case of an alternative series?

Is it the theorem that states that the sum and the integral should both either converge or diverge? ANd can we apply the integral test in the case of an alternative series.

5. Jul 25, 2009

### zcd

For this particular case, $$\frac{1}{n}\sin(\frac{1}{n})$$ is monotonic decreasing and positive for n>1, meaning the series doesn't alternate. This can be verified by noting that its first derivative is always negative.

For alternating series, the integral test cannot be used since it requires the function be positive and always decreasing on an interval $$[a,\infty)$$.

If f(x) is monotonic decreasing and positive, and $$\int_{a}^{\infty}f(x)\,dx$$ converges, then the corresponding series also converges. If the integral diverges, then the series will also diverge.

6. Jul 25, 2009

### Bachelier

Makes sense. Thank you.