Question

HELP PLEASE

Answer 1

Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)

\[a_n = \frac{ 5^{n+2} }{ 6^n }\]

Apparently the answer is 0. But i dont know how to get there.

Answer 2

@zepdrix @518nad @mathmate

Answer 3

\[\large\rm 5^{n+2}=5^n\cdot5^2\]

Answer 4

\[\large\rm a_n=5^2\frac{5^n}{6^n}\]

Answer 5

\[\large\rm a_n=25\left(\frac56\right)^n\]

Answer 6

Hm I got that too. But how does that equal 0?

Answer 7

If you're ever unsure of what a limit is really doing,
just plug in a really large value to get an idea.

(5/6)^10000 = ?

Answer 8

Idk what that is doe. :/

Answer 9

I mean, it's better to understand it intuitively,
but this is a nice shortcut if you ever get stuck on a test,
and are allowed to use a calculator for basic arithmetic

Answer 10

in the form zepdrix wrote, it is essentially a geometric sequence

a_n = a_1 * r^n

which converges to 0 when r < 1

Answer 11

Oh we are allowed to use geometric with sequences? I thought it was only for series

Answer 12

geometric sequence is that really simple thing you learned way back in algebra. I don't know if they reteach it in calculus or not, so maybe it's a bit rusty to you.

Just try to realize that when you square a number which is less than 1,
it gets smaller.

Example: \(\large\rm \left(\frac12\right)^2=\frac14\) smaller than 1/2.

If you raise it to the 3rd power, it gets even smaller.
If you raise it to the bajillionth power, it's essentially 0.

Answer 13

Got it