Simplifying the upper and lower sum of a Riemann Integral

Question

I have been working on solving a sum for a few hours now but I am not getting the answer on my answer sheet (which is $\frac{46}{3}$).

I have two sums:

$16(\sum^n_{i=1} \frac{i^2-2i+1}{n^3}+\frac{i-1}{n^2})+\sum_{i=1}^n\frac{2}{n}$

$16( \sum_{i=1}^{n}\frac{i^2}{n^3}+\frac{i}{n^2})+\sum_{i=1}^n\frac{2}{n}$

both of which should converge to $\frac{46}{3}$ as n approaches infinity.

I know I also need to use this $\sum^n_{i=1} i^2 = \frac{n^3}{3}+ \frac{n^2}{2}+ \frac{n}{6}$ in order to answer the questions.

Could someone please help me do these? I've tried many different approaches all day with no success.

Answer 1

Notice that $n$ has no dependence on $i$.

$$ \sum_{i=1}^n \frac{f(i)}{g(n)} = \frac{1}{g(n)} \sum_{i=1}^n f(i) \text{,} $$ which gets all the $n$s in the denominators out of the sums...

Then $$ \sum_{i=1}^n (a i^2 + b i + c) = a\sum_{i=1}^n i^2 + b \sum_{i=1}^n i + \sum_{i=1}^n c $$ (... since addition is commutative).

Answer 2

Here's how to do one. If you follow it, you should be able to do the other.

$$16\left[ \sum\limits_{i=1}^{n}\frac{i^2}{n^3}+\frac{i}{n^2}\right]+\sum\limits_{i=1}^n\frac{2}{n}$$

$$=16\left[\frac1{n^3}\left(\frac{n^3}3+\frac{n^2}2+\frac n6\right)+\frac1{n^2}\left(\frac{n^2}2+\frac n2\right)\right]+2$$

$$\to 16\left[\frac13+\frac12\right]+2=\dfrac{46}3$$