# Thread: Infinite sums help extremely needed

1. ## Infinite sums help extremely needed

Problem 1 Evaluate

Σ (1/n^2 + 5n + 6)
n=1

And,

Problem 2 Evaluate

Σ (2^2n3^3n/4^4n)
n=1

2. Originally Posted by tapiaghector
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 ...

$\frac{1}{n^2+5n+6} = \frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$

telescoping series ...

$\sum \frac{1}{n+2} - \frac{1}{n+3} = \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + ...$

$\frac{2^{2n} \cdot 3^{3n}}{4^{4n}} = \frac{4^n \cdot 27^n}{256^n} = \left(\frac{27}{64}\right)^n$

$\sum \left(\frac{27}{64}\right)^n$

convergent geometric series, right?

3. Thanks for the help my friend, but how would you evaluate the limit as n goes to infinity in the first one? because it is to supposed to be solved by geometric series...

4. Originally Posted by tapiaghector
Thanks for the help my friend, but how would you evaluate the limit as n goes to infinity in the first one? because it is to supposed to be solved by geometric series...
the first series is not geometric ... it's a telescoping series.

do some research and learn how it works.

5. Oh i see what your point, so the answer is going to be the first integer, since the second one in each evaluation will cancel with the next, minus the last fraction.

limit (1/3) - 1/(n+3) = 1/3
n->infinity

Am I mistaking?

6. that's right.