The Maclaurin series for the arctangent function converges for $−1 < x ≤ 1$ and is given by,
$\arctan x=\lim P_{n}(x)$=$\lim \sum_{i=1}^{n}(-1)^{i+1}$$\frac{x^{2i-1}}{2i-1}$
Use the fact that $\tan π/4 = 1$ to determine the number of n terms of the series that need to be summed to ensure that $|4P_n(1) − π| < 10^{-3}$
I don't really know how to proceed with this question. Any ideas?
In this case you are dealing with an alternating series because it so happens that the series representation of arctan is alternating. Thus we can use the remainder therom for alternating series. This states the remainder, $|Rn| \le a_n+1$. Let $x=10^{-3}.$
We have $|4(a_1+…_+a_n)-\pi$| $\le 4|a_n+1| \le x$
So what is $a_{n+1}$? well the general form, is $a_n$=$(-1)^{n+1}/2n+1 $ , so $a_{n+1}$ (taking account absolute values) is $1/(2(n+1)-1=1/2n+1.$
So we want $|4(a_1+…_+a_n)-\pi$| $\le 4/2n+1 \le x$ = 10^${-3}$
equivelently, we want $(2n+1)/4$ $\ge 1/10^{-3}$ i.e. 2n+1 $\ge 4/10^{-3}$ , and in conclusion, we can see we want $n \ge$ (4/10^${-3}$)-1)/2), or n $\ge 3999/2$ but n is an integer so it has to be $n \ge 2000$
