Show that $\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1$ 
Show that $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1 .$$

I know that $\arctan 1 = \frac{\pi}{4}$ and that the sequence being subtracted is a partial sum of its Taylor series. I believe you use the alternating series test to explain, but all I get from it is that the series will converge on $[-1,1]$.
 A: you can actually get a stronger result such as $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 1/11 $$
because the error in a convergent alternating series is less than the absolute value of the  first missing term which in this case is $1/11.$
The alternating series is simply the Taylor series of $\tan ^{-1} x $ evaluated at $x=1$ and it converges by the alternating series test. 
A: Consider the function 
$$\frac{x^{10}}{x^2+1} \equiv 1-x^2+x^4-x^6+x^8-\frac{1}{x^2+1}$$
The integral:
$$J = \int_0^1 \frac{x^{10}}{x^2+1} \text{d}x = \int_0^1 1-x^2+x^4-x^6+x^8-\frac{1}{x^2+1} \text{d}x$$
is clearly positive.
Since $1\leqslant x^2+1 \leqslant 2$, we have:
$$\frac{1}{2} \int_0^1 x^{10} \text{d}x < J < \int_0^1 x^{10} \text{d}x$$
$$\frac{1}{22} < J < \frac{1}{11}$$
Then since $|-J|=J$ and $\frac{1}{11} < \frac{1}{10}$, the desired inequality follows.
A: Use the series expansion of arctangent and the fact that for alternating series the rest is less than the first omitted term.
A: Simple answers above. The alternative is as follows:


*

*We know that $$1-\frac13+\frac15-\frac17+\frac19=\frac{263}{315}$$ which can be calculated by hand.

*We also know that $$\frac\pi4<\frac{3.2}4=0.8\quad\text{whereas}\quad\frac{263}{315}>\frac{252}{315}=0.8$$ so the inequality is equivalent to $$\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)-\frac{\pi}{4} < 0.1\iff\frac\pi4>\frac{463}{630}\iff\pi>\frac{926}{315}$$

*This is true, since $\pi>3$ and $$\frac{926}{315}<\frac{945}{315}=3.$$
