How would I evaluate:
$$\lim_{n \to \infty}\sum_{i=1}^n\frac{2}{n}\left[ \left( \frac{3i}{n} \right)^3+\frac{5i}{n}+2\right]$$
So far, I have cubed the inner bracket and taken the $\dfrac{27}{n^3}$ outside to get:
$$\lim_{n \to \infty} \sum_{i=1}^n \frac{54}{n^4}\left(i^3+\frac{5n^2i}{27}+\frac{2n^3}{27}\right).$$
Then I would take the summation of each individual term while substituting the summation of $i^3$ from $i=1$ to $n$ as $(n^2(n+1)^2)/4$.
What is confusing me here is the other terms, because they still $n$ and $i$ mixed together, I don't know to to deal with them.
Any help is appreciated.
Best Answer
Of course, this can be massively simplified with a bit of integration...
Consider the function $f:x\mapsto2((3x)^3+5x+2)$, then the $n$th sum is $$ \frac1n\sum_{i=1}^nf\left(\frac{i}n\right), $$that is, the $n$th Riemann sum of $f$ on the interval $(0,1)$. Since $f$ is continuous, the limit exists and is $$ \int_0^1f(x)\,\mathrm dx = \left.2\left(3^3\frac14x^4+5\frac12x^2+2x\right)\right|_0^1=2\left(\frac{27}4+\frac52+2\right)=\frac{45}2. $$