Question

how to find the interval of convergence of tan^-1 x

Answer 1

the maclaurin series of tan^-1 x is:
\[x-\frac{ x ^{3} }{ 3! }+\frac{ x^5 }{5}-\frac{ x^7 }{ 7 }+...\]

Answer 2

I assume you mean the interval of convergence for the power series representation, right?

Answer 3

yes

Answer 4

Okay, so in general, we have that the terms of the series are
\[a_n = \frac{x^{2n+1}}{2n+1} \]
do you agree?

Answer 5

yes

Answer 6

wait, what about the -1^k?

Answer 7

Oops, add that in there, my mistake.

Answer 8

To find the interval of convergence, we will demand that
\[\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| < 1 \]

When we plug in, we get:
\[ \lim_{n\rightarrow \infty} x^2 \left( \frac{2n+1}{2n+3} \right) = x^2 < 1 \]
so the power series is valid for all x such that -1 < x < 1.

Answer 9

where hasthe (2n+1)/(2n+3) gone?

Answer 10

?

Answer 11

\[ a_n = \frac{x^{2n+1}}{(2n+1)!}\]

\[ {a_{n+1} \over a_n} = \frac{x^{2n+3}}{(2n+3)!}\times \frac{(2n+1)!}{x^{2n+1}} = {x^2\over (2n+2)(2n+3)} \]

converges for all values of x.

Answer 12

ok, thx!

Answer 13

No, that's not true. There's no factorial.

Answer 14

@experimentX

Answer 15

@tiffanymak1996 The series converges for |x| < 1 . The fraction you asked about disappears because as you take the limit as n -> infinity, that fraction becomes 1.

Answer 16

hold on ... i;ven't check
looks like you are right
http://www.wolframalpha.com/input/?i=expand+tan^-1+x+at+x%3D0

Answer 17

for x=1