PX153 - F3 - estimating the sum of a series

consider the series $S = \sum_{n - 1}^{\infty} a_{n}$ where $a_{n} > 0$
suppose that $n > N$ , we can write $a_{n} = f (n)$ , where $n = N + 1, N + 2, . . .$
eg: for $S = \sum_{n - 1}^{\infty} \frac{1}{n^{2}}$ , we assume that above some $N$ , we can treat $f (n) = \frac{1}{n^{2}}$
- here, $f (x)$ is a positive, decreasing, and continuous function for $N < x < \infty$
- we then have:
$S - S_{N} = \sum_{n = N + 1}^{\infty} f (n) \leq \int_{N}^{\infty} f (x) . d x \equiv A_{N}$
- integral is the orange area in figure
- similarly, we have
$S - S_{N} \geq \int_{N + 1}^{\infty} f (x) . d x \equiv A_{N + 1}$
- integral is the blue area in figure
- it is translated by -1 in the x direction
- thus, $A_{N + 1} + S_{N} \leq S \leq A_{N} + S_{N}$
$S ≊ S_{N} + \frac{1}{2} (A_{N} + A_{N + 1})$
eg: estimate $S = \sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ using the first five terms
$S_{5} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} = 1.464$
$A_{5} = \int_{5}^{\infty} \frac{1}{x^{2}} . d x = [- \frac{1}{x}]_{5}^{\infty} = \frac{1}{5}$
$A_{6} = \frac{1}{6}$
$⟹ 1.464 + \frac{1}{6} \leq S \leq 1.464 + \frac{1}{5}$
- best guess: $S = 1.464 + \frac{1}{2} (\frac{1}{5} + \frac{1}{6}) = 1.647$
- actual value: $S = \frac{π^{2}}{6} = 1.645$