# Problem finding the limit of the remainder of a series

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

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

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

$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

$
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

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

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

$
\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 $R_n(x)$ as 0?
Book showed and I understand that the limit of $\frac{\left|x\right|^{(2n+3)}}{2n+3}$ is 0 when n
approaches $\infty$. But that's only half of it? The author did not calculate
the limit of integral $R_n(x)$ 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

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

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

$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

$
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

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

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

$
\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 $R_n(x)$ as 0?
Book showed and I understand that the limit of $\frac{\left|x\right|^{(2n+3)}}{2n+3}$ is 0 when n
approaches $\infty$. But that's only half of it? The author did not calculate
the limit of integral $R_n(x)$ 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?

I don't understand your question.

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

CB