Computing the value of a series by telescoping cancellations vs. infinite limit of partial sums $$\sum_{m=5}^\infty \frac{3}{m^2+3m+2}$$
Given this problem my first approach was to take the limit of partial sums. To my surprise this didn't work. Many expletives later I realized it was a telescoping series. 
My question is why my first approach failed. My expectation is that both approaches would produce the same answer. Why didn't they?
First approach:
$$\sum_{m=5}^\infty \frac{3}{m^2+3m+2} = \lim \limits_{N \to \infty} \int_{m=5}^N \frac{3}{m^2+3m+2} = \lim \limits_{N \to \infty} 3 \left [ \ln \left ( \frac{N+1}{N+2} \right ) + \ln \left ( \frac{7}{6} \right ) \right ]$$$$ = 3 \ln(7/6) \approx 0.46245$$

An empirical check showed that the above approach is wrong. After I realized it was a telescoping series I was able to produce a sequence of partial sums:

$$S_{m} = \left ( \frac{3}{6}-\frac{3}{m+2} \right ) $$

And the limit of this sequence gets to an answer that agrees with a crude empirical spreadsheet validation:

$$\lim \limits_{m \to \infty} \left ( \frac{3}{6}-\frac{3}{m+2} \right ) = \frac{1}{2}$$
So clearly my initial intuition and understanding was wrong. But why was it wrong? I thought I could take the limit of an integral to calculate the value of a series. In what case do we apply the first approach I took? I've used it before, but I must have forgotten how to apply it correctly (and all my searches come up with references to computing convergence, not actual values).

You can sometimes use this idea to show a series converges (or diverges) but it will not give you the sum. Think about it: an impoper integral is the limit of a  Riemann sum over an interval $[a,b]$ on a continuous variable $x$,  after which you take a limit $b\rightarrow \infty$. Your "integral" is over a discrete variable (the integers) and makes no sense. You actually found 
$\int_{5}^{\infty}\frac{3dx}{x^{2}+3x+2}$
