# Problem finding the limit of the remainder of a series

• Mar 20th 2010, 01:15 PM
x3bnm
Problem finding the limit of the remainder of a series
Suppose there is a series

$\displaystyle \frac{1}{1+t^2} = 1 -t^2 + t^4 - t^6 + ... + (-1)^n t^{2n} + \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}$

in which the last term comes from adding the remaining terms as a
geometric series with the first term $\displaystyle a = (-1)^{(n+1)} t^{(2n+2)}$
and ratio $\displaystyle r= -t^2$ Integrating both sides of equation from
$\displaystyle t = 0$ to $\displaystyle t = x$ gives

$\displaystyle tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... + (-1)^n \frac{x^{(2n+1)}}{2n+1} + R_n(x)$

where

$\displaystyle R_n(x) = \int_0^x \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}dt$

The denominator of integrand is greater than or equal to 1;hence

$\displaystyle \left|R_n(x)\right| \le \int_0^{\left|x\right|} t^{(2n+2)} dt = \frac{\left|x\right|^{(2n+3)}}{2n+3}$

if $\displaystyle \left|x\right| \le 1$ , the right side of this inequality approaches zero as $\displaystyle n \to \infty$
Therefore

$\displaystyle \lim _{n \to \infty} R_n(x) = 0\,\, if \left|x\right| \le 1$

Now my question is how the book found the limit of $\displaystyle R_n(x)$ as 0?
Book showed and I understand that the limit of $\displaystyle \frac{\left|x\right|^{(2n+3)}}{2n+3}$ is 0 when n
approaches $\displaystyle \infty$. But that's only half of it? The author did not calculate
the limit of integral $\displaystyle R_n(x)$ as a whole. I don't understand how he
reached the conclusion? Obviously I'm missing something simple. Is
there any theorem that i'm unaware of? Anyone knows how
he reached the conclusion?
• Mar 21st 2010, 10:42 AM
CaptainBlack
Quote:

Originally Posted by x3bnm
Suppose there is a series

$\displaystyle \frac{1}{1+t^2} = 1 -t^2 + t^4 - t^6 + ... + (-1)^n t^{2n} + \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}$

in which the last term comes from adding the remaining terms as a
geometric series with the first term $\displaystyle a = (-1)^{(n+1)} t^{(2n+2)}$
and ratio $\displaystyle r= -t^2$ Integrating both sides of equation from
$\displaystyle t = 0$ to $\displaystyle t = x$ gives

$\displaystyle tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... + (-1)^n \frac{x^{(2n+1)}}{2n+1} + R_n(x)$

where

$\displaystyle R_n(x) = \int_0^x \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}dt$

The denominator of integrand is greater than or equal to 1;hence

$\displaystyle \left|R_n(x)\right| \le \int_0^{\left|x\right|} t^{(2n+2)} dt = \frac{\left|x\right|^{(2n+3)}}{2n+3}$

if $\displaystyle \left|x\right| \le 1$ , the right side of this inequality approaches zero as $\displaystyle n \to \infty$
Therefore

$\displaystyle \lim _{n \to \infty} R_n(x) = 0\,\, if \left|x\right| \le 1$

Now my question is how the book found the limit of $\displaystyle R_n(x)$ as 0?
Book showed and I understand that the limit of $\displaystyle \frac{\left|x\right|^{(2n+3)}}{2n+3}$ is 0 when n
approaches $\displaystyle \infty$. But that's only half of it? The author did not calculate
the limit of integral $\displaystyle R_n(x)$ as a whole. I don't understand how he
reached the conclusion? Obviously I'm missing something simple. Is
there any theorem that i'm unaware of? Anyone knows how
he reached the conclusion?

The author showed that the remainder in the series for arctan goes to zero as $\displaystyle n$ becomes arbitrarily large for all $\displaystyle x$ such that $\displaystyle |x|\le 1$, what more do you need?

CB