# Series problem involving trig. function

• Feb 2nd 2008, 05:11 PM
lllll
Series problem involving trig. function
I'm having trouble with 3 series questions involving trig.

1. $\displaystyle \sum_{n=1}^\infty \frac{cos^2 (n)}{n^2 +1}\$

2. $\displaystyle \sum_{n=0}^\infty \frac{1+ sin (n)}{10^n}\$

3. $\displaystyle \sum_{n=1}^\infty \frac{arctan (n)}{n^{1.2}}\$

For the most part I know that they are convergent, but I am unable to show it. I have attempted to demonstrate it using a limit comparison test, but have so far been unsuccessful.
• Feb 2nd 2008, 05:23 PM
Jhevon
Quote:

Originally Posted by lllll
I'm having trouble with 3 series questions involving trig.

1. $\displaystyle \sum_{n=1}^\infty \frac{cos^2 (n)}{n^2 +1}\$

note that $\displaystyle \sum_{n = 1}^{\infty} \frac {\cos^2 n}{n^2 + 1} \le \sum_{n = 1}^{\infty} \frac 1{n^2}$ for all $\displaystyle n \in \mathbb{N}$

use the (direct) comparison test
• Feb 2nd 2008, 05:24 PM
Jhevon
Quote:

Originally Posted by lllll
2. $\displaystyle \sum_{n=0}^\infty \frac{1+ sin (n)}{10^n}\$

similar to the last one: $\displaystyle \sum_{n = 0}^{\infty} \frac {1 + \sin n}{10^n} \le \sum_{n = 0}^{\infty} \frac 2{10^n}$
• Feb 2nd 2008, 05:27 PM
Jhevon
Quote:

Originally Posted by lllll
3. $\displaystyle \sum_{n=1}^\infty \frac{arctan (n)}{n^{1.2}}\$

same thought: $\displaystyle \sum_{n = 1}^{\infty} \frac {\arctan x}{n^{1.2}} \le \frac {\pi}2 \sum_{n = 1}^{\infty} \frac 1{n^{1.2}}$
• Feb 2nd 2008, 06:53 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
similar to the last one: $\displaystyle \sum_{n = 0}^{\infty} \frac {1 + \sin n}{10^n} \le \sum_{n = 0}^{\infty} \frac 2{10^n}$

You should say: $\displaystyle \left| \frac{1+\sin n}{10^n} \right| \leq \frac{2}{10^n}$.

Because the direct comparasion test does not work otherwise.

Or, you can write $\displaystyle 0\leq \frac{1+\sin n}{10^n} \leq \frac{2}{10^n}$. Again, because if you do not include that the test does not necessary work (and it is easy to find an example).