Series problem involving trig. function

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February 2nd 2008, 05:11 PM
lllll

Series problem involving trig. function

I'm having trouble with 3 series questions involving trig.

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

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

3. $\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.
February 2nd 2008, 05:23 PM
Jhevon

Quote:

Originally Posted by lllll

I'm having trouble with 3 series questions involving trig.

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

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

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

Quote:

Originally Posted by lllll

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

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

Quote:

Originally Posted by lllll

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

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

Quote:

Originally Posted by Jhevon

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

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

Because the direct comparasion test does not work otherwise.

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

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