I have encountered this problem:
Part a) is easy.
However the problems arose when I tried to do the part b).
I did it this way:
$S = \sum_{n=1}^{\infty} \frac{1}{n^3}$.
$\sum_{n=1}^{3} \frac{1}{n^3} + \int_{4}^{\infty} \frac{1}{x^3} < S < \sum_{n=1}^{3} \frac{1}{n^3} + \int_{3}^{\infty} \frac{1}{x^3}$
Evaluating the LHS and RHS I get that: $1.193 < S < 1.218$.
However, the solution should be the following:
Could anyone elaborate? I know that both bounds (mine and the markscheme are correct, as the original sum is between them; I do not understand why their solution takes into only the first two rectangles + integral as a lower bound.
Best Answer
By the right hand side of (a):
$$S=1+\frac{1}{2^3}+\left(\frac{1}{3^3}+...\right)>1+\frac{1}{2^3}+\int^{\infty}_3 \frac{1}{x^3}\,dx=1+\frac{1}{8}+\frac{1}{18}$$
By the left hand side of (a):
$$S=1+\frac{1}{2^3}+\frac{1}{3^3}+\left(\frac{1}{4^3}+...\right)<1+\frac{1}{2^3}+\frac{1}{3^3}+\int^{\infty}_3 \frac{1}{x^3}\,dx=1+\frac{1}{8}+\frac{1}{27}+\frac{1}{18}$$
The $\frac{1}{18}$ comes from the evaluation of the integral.
Yours is also correct. It just didn't completely use the result from part (a).
Your lower bound is actually better because you used one more term. You can always use more and more terms to get better bounds. But I think the hence in part (b) emphasizes you use the result from part (a).