Problem source

Prove that 
\[
\tan^{-1}t=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\cdots+\frac{(-1)^{n}t^{2n+1}}{2n+1}+(-1)^{n+1}\int_{0}^{t}\frac{x^{2n+2}}{1+x^{2}}\,\mathrm{d}x.
\]
Hence show that, if $0\leqslant t\leqslant1,$ then 
\[
\frac{t^{2n+3}}{2(2n+3)}\leqslant\left|\tan^{-1}t-\sum_{r=0}^{n}\frac{(-1)^{r}t^{2r+1}}{2r+1}\right|\leqslant\frac{t^{2n+3}}{2n+3}.
\]
Show that, as $n\rightarrow\infty,$ 
\[
4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}\rightarrow\pi,
\]
but that the error in approximating $\pi$ by ${\displaystyle 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}}$
is at least $10^{-2}$ if $n$ is less than or equal to $98$. 

Solution source

We start by noticing that $\displaystyle \tan^{-1} t = \int_0^t \frac{1}{1+x^2} \d x$. 

Consider the geometric series $1-x^2+(-x^2)^2+ \cdots + (-x^2)^n = \frac{1-(-x^2)^{n+1}}{1+x^2}$. Therefore, $(1+x^2)(1-x^2+(-x^2)^2+ \cdots + (-x^2)^n) = 1-(-x^2)^{n+1}$ or

$1 = (1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}$
\begin{align*}
\tan^{-1} t &= \int_0^t \frac{1}{1+x^2} \d x \\
&= \int_0^t \frac{(1+x^2)(1-x^2+x^4-\cdots+(-1)^nx^{2n}) +(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\
&= \int_0^t (1-x^2+x^4-\cdots+(-1)^nx^{2n})\d x + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\
&= t - \frac{t^3}{3}+\frac{t^5}{5}-\cdots + (-1)^n \frac{t^{2n+1}}{2n+1}+\int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\
&= \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} + \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \\
\end{align*}

Therefore we can say (for $0 \leq t \leq 1$)

\begin{align*}
\left | \tan^{-1} t -  \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\
&\leq \left | \int_0^t x^{2n+2} \d x \right | \\
&= \frac{t^{2n+3}}{2n+3} \\
\\
\left | \tan^{-1} t -  \sum_{r=0}^n \frac{(-1)^r t^{2r+1}}{2r+1} \right | &= \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{x^2+1} \d x \right | \\
&\geq  \left | \int_0^t \frac{(-1)^{n+1}x^{2n+2}}{1+1} \d x \right | \\
&= \frac{t^{2n+3}}{2(2n+3)} \\
\end{align*}

Since $\tan^{-1} 1 = \frac{\pi}{4}$ we must have that:

\begin{align*}
\lim_{n \to \infty} \left | \frac{\pi}{4} - \sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)} \right | \to 0 \Rightarrow \lim_{n \to \infty}  4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}  \to \pi
\end{align*}

However, 

\begin{align*}
&& \left | 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}  - \pi \right | &\geq 4 \frac{1}{2(2n+3)} \\
&& &= \frac{2}{2n+3} \\
\\
&& \frac{2}{2n+3} \geq 10^{-2} \\
\Leftrightarrow && 200 \geq 2n+3 \\
\Leftrightarrow && 197 \geq 2n \\
\Leftrightarrow && 98.5 \geq n \\
\end{align*}

Therefore we need more than $98$ terms to get two decimal places of accuracy. Not great!