No issues with convergence here. If you actually expand the summand you can use the well-known Faulhaber formulas to compute the sum.
[tex]\displaystyle \sum_{i=1}^n 1 = n[/tex]
[tex]\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}2[/tex]
[tex]\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}6[/tex]
[tex]\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}4[/tex]
and so on. You would end up with
[tex]\displaystyle L_n = \frac2n \times \frac{96-8796n^2+40283n^3-49458n^4}{n^3}[/tex]
As [tex]n\to\infty[/tex], we have [tex]L_n\to-98916[/tex].
Now, when [tex]i=1[/tex], the first term of the sum is
[tex]7(5+0)^3 - 6(5+0)^5[/tex]
so the left endpoint of the first subinterval is [tex]a=5[/tex].
At the other end, when [tex]i=n[/tex], the last term of the sum is
[tex]7\left(5+\dfrac{2(n-1)}n\right)^3 - 6\left(5+\dfrac{2(n-1)}n\right)^5[/tex]
which converges to
[tex]7(5+2)^3 - 6(5+2)^5[/tex]
so the left endpoint of the last subinterval converges to [tex]b=5 + 2 = 7[/tex].
From here it's quite clear that [tex]f(x)=7x^3-6x^5[/tex]. So, the Riemann sum converges to the definite integral
[tex]\boxed{\displaystyle \int_5^7 (7x^3 - 6x^5) \, dx}[/tex]