Question

for which values of x this converge http://puu.sh/fIjqD/964b4a42c9.png

Answer 1

I think the ratio test might help

Answer 2

i tried that, no help, can u try and see?

Answer 3

\[\text{ Let } L=\lim_{n \rightarrow \infty}|\frac{a_{n+1}}{a_n}| \\ \text{ if } L<1 \text{ then } a_n \text{ converges } \\ \text{ if } L>1 \text{ then } a_n \text{ diverges } \\ \text{ if } L=1 \text{ test is inconclusive }\]

Answer 4

\[L=\lim_{n \rightarrow \infty}|\frac{(n+1)! (10x-11)^{n+1}}{11^{n+1}} \cdot \frac{11^n}{n! (10x-11)^n }|\]

Answer 5

\[L=\lim_{n \rightarrow \infty}|\frac{(n+1)! (10x-11)^{n+1}}{11^{n+1}} \cdot \frac{11^n}{n! (10x-11)^n }| \\ L=\lim_{n \rightarrow \infty}|\frac{(n+1) \cdot n! \cdot (10x-11)^n \cdot (10x-11) \cdot 11^n }{11 ^n \cdot 11 \cdot n! \cdot (10x-11)^n}|\]
you should see alot of cancellation there

Answer 6

are you there?
like is that the part you are stuck on

Answer 7

yea i'm here, i'm stuck here because this gives me
lim n->infinity of 1/11 * (10x-11) * (n+1)!/n! =
lim n->infinity of 1/11 * (10x-11) * infinity

what do i do next?

Answer 8

@freckles

Answer 9

(n+1)!=(n+1)*n!
so (n+1)!/n!=(n+1)

Answer 10

so we should be looking at
\[\lim_{n \rightarrow \infty}|\frac{(n+1)(10x-11)}{11}|\]

Answer 11

LOL ye my bad im retard

Answer 12

for it doesn't matter what x is
(n+1) is going to go to infinity right?

Answer 13

even if we consider 10x-11=0
which is when x=11/10
and run the ratio test again
we will still get a similar result

Answer 14

\[\text{ if } x=\frac{11}{10} \text{ we have } a_n=\frac{n!(0)^n}{11^n} =\frac{n!}{11^n}\]
and
\[\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{11^{n+1}} \cdot \frac{11^n}{n!}=\frac{n+1}{11}\]

Answer 15

and still goes to infinity when n goes to infinity

Answer 16

so it seems there is value x such that the series converges

Answer 17

thanks i get what u mean, its 11/10

Answer 18

well I was saying it still diverges then

Answer 19

i meant to say it seems there is No value x

Answer 20

such that the series converges

Answer 21

i think there is 11/10, the only one

Answer 22

but if x=11/10 isn't a_n=n!/11^n

Answer 23

\[\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{11^{n+1}} \cdot \frac{11^n}{n!}=\frac{n+1}{11}->\infty \text{ as } n->\infty \]

Answer 24

\[L>1 \text{ so } a_n \text{ diverges when } x=\frac{11}{10}\]

Answer 25

so for all n we have L>1

Answer 26

oops I mean the series diverges