# Homework Help: Calculating a limit of a sequence

1. Jan 22, 2012

### lazy.man

Hi guys,

1. The problem statement, all variables and given/known data
I'm trying to calculate the limit of the following sequence:
an:= ($\frac{n^{2}+10}{n^{2}-5n}$)$^{(3n^{2}+2)}$​

2. The attempt at a solution
Ok so I got this so far:

an = $\left(\frac{\left(n^2\left(1+\frac{10}{n^2}\right)\right)}{\left(n^2\left(1-\frac{5}{n}\right)\right)}\right)^{n^{2}(3+\frac{2}{n})}$​

$\lim_{n\to \infty }$ an = $\lim_{n\to \infty }$ 1$^{∞}$​

So this is what I got till now and I don't know really how to proceed or how can I approach this problem differently to find the correct solution which is ∞.

Any help, tips are welcome.

2. Jan 22, 2012

### SammyS

Staff Emeritus
Without having worked the problem out, my first thought is to find the limit of the log of this.

If $\displaystyle \lim_{n\to\infty}\ln(a_n)=L\,,\ \text{ then } \lim_{n\to\infty}a_n=e^L$

3. Jan 23, 2012

### dirk_mec1

I don't know if this is allowed with products of limits rule but:

$$3n^2 \ln \left( \frac{n^2}{n-5n}\right) = 3n \ln \left[\left( 1 +\frac{5}{n} \right)^n \right] = 3n \cdot 5 = 15n \rightarrow \infty$$

4. Jan 23, 2012

### Curious3141

Binomial theorem.

${(\frac{n^{2}+10}{n^{2}-5n})}^{(3n^{2}+2)} = {(1 + \frac{5}{n} + \frac{35}{n(n-5)})}^{(3n^2 + 2)} > {(1 + \frac{5}{n})^{3n^2} = {[{(1+\frac{5}{n})}^n}]}^{3n}$, the limit of which is $e^{15n}$, which is ∞.

This is a slight (more direct) variation of dirk_mec's idea.

EDIT: BTW, I messed up the sign, clearly it should be greater than ('>').

Last edited: Jan 23, 2012
5. Jan 23, 2012

### lazy.man

Thank you guys for your help and I'm posting my solution to the problem.