# Upper and Lower Integral Bounds on Infinite Sum

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.

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).

• thanks, but what is the difference between my part and their part? Since I was taught to approximate the infinite sum using my method above. – jermenkoo May 16 '15 at 10:57
• @jermenkoo: Come to think about it, yours is also correct. It just didn't use the result from part (a). You used one further step instead. – KittyL May 16 '15 at 11:00
• Shouldn't the integral in the LHS part be: $\int^{\infty}_4 \frac{1}{x^3}\,dx$ – mavavilj Sep 17 '15 at 17:23
• @mavavilj: It is the right hand sum of $\int_3^{\infty} \frac{1}{x^3} dx$, that's why it is "$<$". – KittyL Sep 17 '15 at 23:09
• @KittyL Ah correct. Had misread the integral test proof bit, but the extra term is seen there as well. – mavavilj Sep 18 '15 at 17:14