[Solved] Evaluate the limit by recognizing the Riemann sum

Evaluate $\displaystyle \lim_{n\to\infty}\frac{1+2+3+...+n}{n^2+n}$ by first recognizing the sum as a Riemann sum over a partition of [0,1] and then evaluating the corresponding integral.A. 0

B. 1/2

C. 1/4

D. Not enough information

E. None of the above

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What is confusing me about the question is whether I'm being asked to compute a Riemann sum or being asked to "reverse engineer" the given function into a Riemann sum. I tend to lean towards the latter so this is what I've done so far...

Given:

$\displaystyle \displaystyle\int^b_a f\left(x\right)\,dx = \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n}f\left (x_i^*\right)\Delta x$

$\displaystyle \Delta x=\frac{(b-a)}{n}$

$\displaystyle x_i^*=a+(\Delta x)i$

Then:

$\displaystyle \Delta x=\frac{\left(1-0\right)}{n}=\frac{1}{n}$

$\displaystyle \frac{1+2+3+...+n}{n^2+n}=\displaystyle\sum_{i=1}^ {n}\frac{i}{n^2+n}=\displaystyle\sum_{i=1}^{n}\fra c{i}{n+1}\cdot\frac{1}{n}$

When I decompose the sum into $\displaystyle \displaystyle\sum_{i=1}^{n}\frac{i}{n+1}\cdot\frac {1}{n}$, I think I'm seing a basis of a Riemann sum with the following:

$\displaystyle f\left(x_i^*\right)=\frac{i}{n+1}$

$\displaystyle \Delta x=\frac{1}{n}$

If I'm correct, then I continue...

$\displaystyle \displaystyle\sum_{i=1}^{n}\frac{i}{n+1}\cdot\frac {1}{n}=\frac{1}{n}\left(\frac{1}{n+1}\cdot\display style\sum_{i=1}^{n}i\right)=\frac{1}{n}\left(\frac {1}{n+1}\cdot\frac{n(n+1)}{2}\right)=\frac{1}{n}\c dot\frac{n}{2}=\frac{1}{2}$

Assuming all of that is correct, the lim as n approaches infinity is 1/2. However, I've spent a lot of time looking at this problem and might be completely off track.

Thanks,

Manny