How to calculate $S_N=\sum\limits_{n=1}^{N}\ln\left(1+\frac{2}{n(n+3)}\right)$? 
How to calculate $S_N=\sum\limits_{n=1}^{N}\ln\left(1+\frac{2}{n(n+3)}\right)$?

I said it was equivalent to:
$$\Leftrightarrow=\sum\limits_{n=1}^{N} \ln(n+1)+\sum\limits_{n=1}^{N} \ln(n+2)+\sum\limits_{n=1}^{N} -\ln(n)-\sum\limits_{n=1}^{N} \ln(n+3)$$
I'm guessing about Taylor developement, but I'm not skilled enough to know how to apply them...
 A: $$\sum_1^N\ln(n+1)+\sum_1^N\ln(n+2)-\sum_1^N\ln(n)-\sum_1^N\ln(n+3) \\
= \sum_2^{N+1}\ln(n)+\sum_3^{N+2}\ln(n)-\sum_1^N\ln(n)-\sum_4^{N+3}\ln(n) \\
 = \ln(N+1)+\ln(3)-\ln(N+3)$$
So $$S_N \to \ln(3)$$
A: Since $\ln(ab) = \ln(a) + \ln(b)$
$$S_N=\sum\limits_{n=1}^{N}\ln\left(1+\frac{2}{n(n+3)}\right) = \ln\left(\prod_{n=1}^N\frac{n^2 + 3n + 2}{n(n+3)} \right) = \ln\left(\prod_{n=1}^N\frac{(n+1)(n+2)}{n(n+3)} \right)$$
$$=\ln\left(\frac{6}{2}\frac{(N+1)!(N+2)!}{N!(N+3)!} \right) = \ln\left(3\frac{N+1}{N+3} \right) $$
A: Use the property $\ln(a)+\ln(b)=\ln(ab)$ to rewrite $\sum \ln$ as $\ln\prod$. If you write out the first few terms of the product (keep the factorizations, don't multiply out) you should notice a good amount of cancellation. Indeed, all but a few identifiable factors will cancel. See if you can identify which factors will survive the mass cancellation.
A: Express each term as a ratio of ratios:
$$1+\frac{2}{n(n+3)} = \frac{(n+1)(n+2)}{n (n+3)} = \frac{\frac{n+2}{n+3}}{\frac{n}{n+1}} $$
As one may see, there is cancellation in every other term.  Thus, the last two terms in the numerator and the first two terms in the denominator survive.  Thus the sum is
$$\log{\left (3 \frac{N+1}{N+3} \right )} $$
