Problem 1 Evaluate
∞
Σ (1/n^2 + 5n + 6)
n=1
And,
Problem 2 Evaluate
∞
Σ (2^2n3^3n/4^4n)
n=1
using partial fractions ...
$\displaystyle \frac{1}{n^2+5n+6} = \frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$
telescoping series ...
$\displaystyle \sum \frac{1}{n+2} - \frac{1}{n+3} = \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + ... $
$\displaystyle \frac{2^{2n} \cdot 3^{3n}}{4^{4n}} = \frac{4^n \cdot 27^n}{256^n} = \left(\frac{27}{64}\right)^n$
$\displaystyle \sum \left(\frac{27}{64}\right)^n$
convergent geometric series, right?