Question

Isn't the series here divergent? doesn't that mean it cannot be represented as a riemann sum

Answer 1

No issues with convergence here. If you actually expand the summand you can use the well-known Faulhaber formulas to compute the sum.

$\displaystyle \sum_{i=1}^n 1 = n$

$\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2$

$\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}6$

$\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}4$

and so on. You would end up with

$\displaystyle L_n = \frac2n \times \frac{96-8796n^2+40283n^3-49458n^4}{n^3}$

As $n\to\infty$, we have $L_n\to-98916$.

Now, when $i=1$, the first term of the sum is

$7(5+0)^3 - 6(5+0)^5$

so the left endpoint of the first subinterval is $a=5$.

At the other end, when $i=n$, the last term of the sum is

$7\left(5+\dfrac{2(n-1)}n\right)^3 - 6\left(5+\dfrac{2(n-1)}n\right)^5$

which converges to

$7(5+2)^3 - 6(5+2)^5$

so the left endpoint of the last subinterval converges to $b=5 + 2 = 7$.

From here it's quite clear that $f(x)=7x^3-6x^5$. So, the Riemann sum converges to the definite integral

$\boxed{\displaystyle \int_5^7 (7x^3 - 6x^5) \, dx}$