Question

alternate series.

Answer 1

I did the ratio test and got a value of zero. My book says it diverges but i don't understand!

Answer 2

\[\large \frac{3^{k+1}k^2}{3^k(k+1)^{2}}\] is a start right?

Answer 3

yes!

Answer 4

okay before we begin and do a bunch of boring algebra, what do you think of
\[\frac{3^k}{k^2}\] you think the terms get bigger or smaller?

Answer 5

smaller?!

Answer 6

hmmm
lets pick a not too big number say 100
now \(100^2\) is pretty big, it is \(10000\) but how about \(3^{100}\)?

Answer 7

well that would be a bigger number ?

Answer 8

yeah, it has maybe 47 zeros

Answer 9

the point is that exponentials, at least with a base larger than one, grows way way faster than a square
in fact they grow faster than any polynomial

Answer 10

so before you begin the algebra, you should expect it to diverge for sure
the terms do not even go to zero! so no way the sum will converge
not only do they get larger, they get larger real fast

Answer 11

that doesn't mean we can't do the algebra
\[\large \frac{3^{k+1}k^2}{3^k(k+1)^{2}}=\frac{3k^2}{(k+1)^2}\] is a start then take the limit as \(k\to \infty\) and what do you get?

Answer 12

aw man i see what you mean. infinity?

Answer 13

no not infinity
\[\lim_{n\to \infty}\frac{3k^2}{(k+1)^2}\] you should do in your head
do you know how?

Answer 14

subscript should have been a "k" obviously
do you know how to do that mentally? if not let me know and i will tell you, it is very easy

Answer 15

im not sure to be honest :/ Id like to hear your advice!

Answer 16

i want to say zero?

Answer 17

ok you are thinking too much
remember in some pre-calc class you learned about horizontal asymptotes? if the degrees of the numerator and denominator of your rational function are the same, like for example
\[\frac{3x^2}{x^2+5x+2}\] then the horizontal asymptote is the ratio of the leading coefficients ?

Answer 18

it is exactly the same here
you want
\[\lim_{k\to \infty}\frac{3k^2}{(k+1)^2}\] the top and bottom are both polynomials of degree 2
the leading coefficient of the numerator is 3
the leading coefficient of the denominator is 1 (which you can see even without multiplying out)
so the limit is \(\frac{3}{1}=3\)

Answer 19

you can do this in your head right?
\[\lim_{k\to \infty }\frac{5k^2}{2k^2+3k}=\frac{5}{2}\]

Answer 20

^ yes! all that slipped my mind |:

Answer 21

yeah because it was in some other class, but it does make it real easy
they try to make it harder i calc by using l'hopital or dividing by the highest power or whatever, but you can do it with your eyeballs

Answer 22

in any case to finish up
the limit is 3 so by the ratio test it diverges
but as the terms do not even go to zero, the ratio test is more than you really need

Answer 23

hmmm i understand :S just need more practice!

Answer 24

yeah of course
good luck!

Answer 25

thank you!