finding the limit for convergence

**mmattson07** · October 27th 2009, 11:08 AM

I can't seem to figure out how this sequence converges to a value.

$\lim_{n\to\infty}\frac{4^{n+2}}{5^n}$

I tried simplifying it to this:

$\lim_{n\to\infty}\frac{(16)4^{n}}{5^n}=$
$16\lim_{n\to\infty}(\frac{4}{5})^n$

But i dont know if that helps it still seems like it diverges, but it doesn't.

**Plato** · October 27th 2009, 11:20 AM

If $|r|<1$ then $\left(r^n\right)\to 0$ .

**tonio** · October 27th 2009, 11:21 AM

Originally Posted by mmattson07

I can't seem to figure out how this sequence converges to a value.

$\lim_{n\to\infty}\frac{4^{n+2}}{5^n}$

I tried simplifying it to this:

$\lim_{n\to\infty}\frac{(16)4^{n}}{5^n}=$
$16\lim_{n\to\infty}(\frac{4}{5})^n$

But i dont know if that helps it still seems like it diverges, but it doesn't.

Very good! Now you only need to remember from H.S. that the limit of an infinite geometric sequence whose quotient is between -1 and 1 is zero and you're done.

Tonio

**mmattson07** · October 27th 2009, 11:27 AM

ah yes thanks guys!

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