[SOLVED] The difference of the harmonic series and 1/n comparsion

**RockHard** · November 18th 2009, 10:22 PM

I recently had this classic example to determine if it converged or diverged
[MATH\frac{sin(n)}{n}][/tex]

I learned that you can use a comparison test using the squeeze theorem.

$\frac{-1}{n}\le\frac{sin(n)}{n}\le\frac{1}{n}$

Then you would take the limit of

$\lim_{n\to\infty}\frac{1}{n}$ which was go to 0.

Because the function was greater than the original you can say the original converged as well to a limit of 0.

However, how would the harmonic series, diverge if you are basically determining the limit of the same thing

Edit: Poor question, trying to mark as solved, please disregard, if misused the concept of series and sequence

**chisigma** · November 18th 2009, 10:47 PM

In...

http://www.mathhelpforum.com/math-he...eries-4-a.html

... it has been demonstrated that the series...

$\sum_{n=1}^{\infty} \frac{\sin n}{n}$ , $\sum_{n=1}^{\infty} \frac{\cos n}{n}$ (1)

... are both convergent and also their sum has been computed. I don't know if it exists an easier way to arrive at the same results... probably it exists