Question

Starting from the geometric series, use power series operations to
determine the Maclaurin series. See picture

Answer 1

a. By replacing $x$ with $-x^2$ in the power series, we get

$\displaystyle \frac1{1+x^2} = \sum_{n=0}^\infty (-x^2)^n = \sum_{n=0}^\infty (-1)^n x^{2n}$

Integrate both sides to recover $\arctan(x)$ on the left.

$\displaystyle \int \frac{dx}{1+x^2} = \int \sum_{n=0}^\infty (-x^2)^n = \sum_{n=0}^\infty (-1)^n x^{2n} \, dx$

$\displaystyle \arctan(x) = C + \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1} \, dx$

By letting $x=0$ on both sides, we find $C=0$, so that

$\displaystyle \arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1} \, dx$

Then dividing both sides by $x$ gives

$\displaystyle \boxed{\frac{\arctan(x)}x = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n} \, dx}$

b. Let $x=\frac1{\sqrt3}$. Then

$\displaystyle \frac{\arctan\left(\frac1{\sqrt3}\right)}{\frac1{\sqrt3}} = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \left(\dfrac1\sqrt3\right)^{2n}$

$\displaystyle \sqrt3 \arctan\left(\frac1{\sqrt3}\right) = \sum_{n=0}^\infty \frac1{2n+1} \left(-\frac13\right)^n$

Taking the first few terms from the infinite series, we can approximate

$n=0 \implies \sqrt3\arctan\left(\dfrac1{\sqrt3}\right) \approx 1$

$0\le n\le1 \implies \sqrt3\arctan\left(\dfrac1{\sqrt3}\right) \approx 1+\dfrac13\left(-\dfrac13\right) = \dfrac89$

which together suggest the value we want is bounded between 8/9 and 9/9 = 1, hence $\boxed{p=8}$.

Since the series is alternating and converges on $-1<x<0\cup0<x<1$,

$\displaystyle \left| \sum_{n=0}^\infty a_n - \sum_{n=0}^0 a_n \right| < |a_1| = \left|\frac13 \left(-\frac13\right)^1\right| = \frac19$

and

$\displaystyle \left| \sum_{n=0}^\infty a_n - \sum_{n=0}^1 a_n \right| < |a_2| = \left|\frac15 \left(-\frac13\right)^2\right| = \frac1{45}$

which tells us the first approximation is off by at most 1/9 from the actual value of $\frac\pi{2\sqrt3}$, whereas the second approximation is off by at most 1/45 from the actual value. In other words, the second approximation is closer, so $\frac\pi{2\sqrt3}$ is closer to $\frac p9$ :

$\dfrac89 \approx 0.8889$

$\dfrac{\pi}{2\sqrt3}\approx0.9069$

$\dfrac99 = 1$