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

Use the series expansion of arctangent and the fact that for alternating series the rest is less than the first omitted term.

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.

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.