# Find upper and lower bounds for the finite sum

Find upper and lower bound for the following finite sum:

$$\frac{1}{1} + \frac{1}{2^3} + \frac{1}{3^3} + ··· + \frac{1}{n^3}$$

My attempt is: Using the integral test:

we know that $$\frac{1}{1} + \frac{1}{2^3} + \frac{1}{3^3} + ··· + \frac{1}{n^3}$$ = $$\sum_{i=1}^n 1/i^3$$ = $$\int_1^n$$1/$$i^3$$di = $$\int_1^n1/x^3$$dx = $$-1/2n^2$$ + $$1/2$$

But now I'm stuck. How can this test give the lower and upper bouunds? Any help please?

The picture below shows how to do it. Best to separate off the initial $$1$$ and take the sum from $$a=2$$ to $$b=n,$$ because the upper bound direction needs an integral beginning at $$a-1.$$
typeset, given $$f > 0$$ and $$f' < 0,$$ we get $$\int_a^{b+1} \; f(x) \; dx \; < \sum_{j=a}^b \; f(j) < \int_{a-1}^b \; f(x) \; dx$$