Answer
A: $S_{10}\approx 1.549$, with an error of $\approx0.096$.
B: $\Sigma^{\infty}_{n=1}\frac{1}{n^{2}}\approx1.644$
C: Difference of $0.000934$ between b estimate and actual value
D: $n>1000$ To get accurate estimation to 3 decimal points
Work Step by Step
Since $\frac{1}{x^2}$ is continuous and decreasing over $(1,\infty)$, the Integral Test can be applied.
A: $ S_{10}=\Sigma^{10}_{n=1}\frac{1}{n^{2}}\approx1.549$, with an error of $\approx0.096$.
B:
Before we solve the inequality we first take the integral and get the limit
$\int_{n}^{\infty}\frac{1}{x^2}dx=$
$\lim\limits_{t \to \infty}\int^{t}_{n}\frac{1}{x^2}dx=$
$\lim\limits_{t \to \infty}[-\frac{1}{x}]_{n}^{t}=\frac{1}{n}$
Now that we have the value, we can now find the inequality
$ S_{n}+\frac{1}{n+1}\leq S \leq S_{n}+\frac{1}{n}$
$1.549+\frac{1}{11}\leq S \leq 1.549+\frac{1}{10}$
$1.6399 \leq S \leq 1.649$
$\frac{1.6399+1.649}{2}=S\approx 1.644$
C:
The difference between the original value $(\frac{\pi^2}{6}) = \frac{\pi^2}{6}-1.644 \approx0.000934$
D: Since we already found the integral in part B, we can reuse it here for the inequality
$\frac{1}{n}\lt1000$
$n>1000$
