# Does a_n=(8000n)/(.0001n^2) converge? If so what is the limit?

Dec 31, 2016

The sequence converges to 0.

#### Explanation:

We can evaluate the limit of this series by looking at dominant terms. The dominant term in the numerator is $n$, while the dominant term in the denominator is ${n}^{2}$.

${\lim}_{n \to \infty} \frac{8000 n}{0.0001 {n}^{2}}$

$\implies {\lim}_{n \to \infty} \frac{n}{n} ^ 2$

$\implies {\lim}_{n \to \infty} \frac{1}{n}$

As $n$ approaches infinity, $\frac{1}{n}$ approaches $0$.

$\implies {\lim}_{n \to \infty} \frac{1}{n} = 0$

$\therefore$ The sequence converges to 0.

Alternatively, you can use L'Hospital's rule to simplify, though this takes a bit more work. It will yield the same result. By this rule, if:

$\lim \frac{f \left(x\right)}{g \left(x\right)} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$, then $\lim \frac{f \left(x\right)}{g \left(x\right)} = \lim \frac{f ' \left(x\right)}{g ' \left(x\right)}$

${\lim}_{n \to \infty} \frac{8000 n}{0.0001 {n}^{2}} \implies \frac{\infty}{\infty}$

Take the derivative of the numerator and denominator:

${\lim}_{n \to \infty} = \frac{8000}{0.0002 n}$

As $n$ approaches infinity, $\frac{8000}{0.0002 n}$ approaches $0$. Essentially, you are dividing $8000$ by bigger and bigger numbers until the number in the denominator is so large that the quotient is $\approx 0$.