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?
On
You do not know that $\sum_{i=1}^n 1/i^3= \int_1^n1/x^3\, dx$ as the equality is not exact. In fact $$\int\limits _1^{n+1}\frac{1}{x^3}\, dx \le \sum_{i=1}^n \frac{1}{i^3}\le 1+ \int\limits_1^n \frac{1}{x^3}\, dx $$
and using your earlier analysis, this becomes $$\frac12 - \frac{1}{2(n+1)^2} \le \sum\limits _{i=1}^n \frac{1}{i^3}\le \frac32 - \frac{1}{2n^2}.$$
This lower bound is not good, as clearly $\sum\limits_{i=1}^n \frac{1}{i^3}\ge 1$ when $n\ge 1$. As Will Jagy suggested, it might therefore be better to use as the lower bound $1+\int\limits _2^{n+1}\frac{1}{x^3}\, dx =$ so $$ \frac{9}{8}-\frac{1}{2(n+1)^2}\le \sum\limits _{i=1}^n \frac{1}{i^3}\le \frac32 - \frac{1}{2n^2}.$$
These bounds are not good for $n\ge 2$ and you could continue the process. An alternative approach might be to start at the other end and to note that the infinite sum is $\zeta(3)= \sum\limits_{i=1}^\infty \frac{1}{i^3} \approx 1.2020569$, known as Apéry's constant, and then subtract $\int\limits _n^{\infty}\frac{1}{x^3}\, dx$ or $\int\limits _{n+1}^{\infty}\frac{1}{x^3}\, dx$ so $$\zeta(3) -\frac{1}{2n^2} \le \sum\limits _{i=1}^n \frac{1}{i^3}\le \zeta(3) -\frac{1}{2(n+1)^2} $$ and these are tighter than the original suggestion for $n\ge 2$ and much tighter as $n$ increases.
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 $$ \int_a^{b+1} \; f(x) \; dx \; < \sum_{j=a}^b \; f(j) < \int_{a-1}^b \; f(x) \; dx$$