1. ## converge/diverge

sum n=1 oo (arctan(n))^2 / 1+n^2

how do i solve it using integral test?

2. Originally Posted by jeph
sum n=1 oo (arctan(n))^2 / 1+n^2

how do i solve it using integral test?
Here

3. Originally Posted by jeph
sum n=1 oo (arctan(n))^2 / 1+n^2

how do i solve it using integral test?
Less formally than Jhevon, as $\arctan(n)$ by definition is giving a value in the range $(-\pi/2, \pi/2)$ $\arctan(n)$ is bounded (in fact not only is it bounded but as $n \to \infty \arctan(n) \to \pi/2$). Also for large $n$, $1/(1+n^2)$ behaves like $1/n^2$ so the series is absolutly convergent.

RonL

4. Originally Posted by CaptainBlack
Less formally than Jhevon, as $\arctan(n)$ by definition is giving a value in the range $(-\pi/2, \pi/2)$ $\arctan(n)$ is bounded (in fact not only is it bounded but as $n \to \infty \arctan(n) \to \pi/2$). Also for large $n$, $1/(1+n^2)$ behaves like $1/n^2$ so the series is absolutly convergent.

RonL
which test is being used right here? o.o

5. Originally Posted by jeph
which test is being used right here? o.o
the comparison test. he wasn't being very formal though, since i already answered the question.

The comparison test says:

Let $\sum a_{n}$ be a series where $a_{n} \geq 0$ for all $n$.

(i) If $\sum a_{n}$ converges and | $b_{n}$| $\leq a_{n}$ for all $n$, then $\sum b_{n}$ converges

(ii) If $\sum a_{n} = + \infty$ and $b_{n} \geq a_{n}$ for all $n$, then $\sum b_{n} = + \infty$.

CaptainBlack used the fact that:

$\sum \left| \frac {\arctan^2 n}{1 + n^2} \right| \leq \sum \frac {(\frac {\pi}{2})^2}{1 + n^2} = \frac {\pi^2}{4} \sum \frac {1}{1 + n^2}$ and $\sum \frac {1}{1 + n^2}$ converges by the comparison test, if we take $\sum a_{n}$ to be $\sum \frac {1}{n^2}$ and $\sum b_{n}$ to be $\sum \frac {1}{1 + n^2}$

EDIT: Bare with me, i'm just learning LaTex, it will take me a while to get used to it
EDIT 2: Finally! It is ready!
EDIT 3: I'm so proud of myself ...even though typing that took forever. i guess i'll get faster with practice though, and after i memorize the commands

6. Originally Posted by Jhevon
i guess i'll get faster with practice though, and after i memorize the commands
It takes just seconds for me. As fast as I type.

7. Originally Posted by ThePerfectHacker
It takes just seconds for me. As fast as I type.
my typing speed is ok...not great, but ok. i am slow because i don't know the commands. i have a LaTex tutorial up while i'm typing and i have to be refering to it constantly. i'll get a lot quicker when i know the commands

8. Originally Posted by jeph
which test is being used right here? o.o
None explicitly, but knowledge of the behaviour of series. This is an informal
argument of the form that we actually use if we want to know if this thing
converges, only later will a formal proof be constructed to make the argument rigorous.

The purpose of such an argument is to explain why this converges, which is not always
clear from a formal proof (Sound of Bourbaki turning in his grave from off stage right).

Jhevon has explained what the idea is in more detail

RonL