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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?

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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 $$ \int_a^{b+1} \; f(x) \; dx \; < \sum_{j=a}^b \; f(j) < \int_{a-1}^b \; f(x) \; dx$$

enter image description here

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