$Arctan(0,2)$ can be represented by $$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$ at x=0,2.Calculate $arctan(0,2)$ with an error less than $5.10^{-5}$.
In this question,I have done the necessary operations and obtained the error(remainder) as $$\frac{(0,2)^{2n+3}}{2n+3}$$.But to show that this is less than $5.10^{-5}$ , I must take n as 2.So,why is n taken as 2?
$$\tan^{-1}(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$ is alternating which means that the error is lest than the first neglected terms.
So, summing up to $m$, the remainder is, in absolute value, $$R_m=\frac{x^{2m+1}}{2m+1}$$ and you want it to be less than $\epsilon$. There is a rigorous solution for $m$ of equation $$\frac{x^{2m+1}}{2m+1}=\epsilon$$ It is given in terms of Lambert function $$m=-\frac{W\left(-\frac{\log (x)}{\epsilon }\right)}{2 \log (x)}-\frac{1}{2}$$ In your case, the argument of $W(z)$ is $z=20000 \log (5)$ which is quite large; so we can approximate the result using the exapnsion given in the Wikipedia page $$W(z)=L_1-L_2+\frac{L_2}{L_1}+\cdots\qquad \qquad (L_1=\log(z)\qquad L_2=\log(L_1))$$ which makes $W(20000 \log (5))\approx 8.265$ and then $m=2.067$.
So, you need three terms to be added.
Let us check $$\left( \begin{array}{ccc} m & R_m & \approx \\ 1 & \frac{1}{375} & 0.00266667 \\ 2 & \frac{1}{15625} & 0.00006400 \\ 3 & \frac{1}{546875} & 0.00000183 \\ 4 & \frac{1}{17578125} & 0.00000006 \end{array} \right)$$