Question

Convergence/Divergence

Answer 1

I have no idea how to approach this.

Answer 2

I mean I get I test for absolute convergence because it says the absolute value of the error but I have no idea otherwise.

Answer 3

Anytime that you have a factorial, I suggest using ratio test.

Answer 4

Solve \(\dfrac{1}{9^{n+1}(n+1)!} < 0.000005\)

Answer 5

According to ratio test, if L < 1, then series is absolutely convergent.

Answer 6

Ahh, That's what I was thinking too. That looks disgusting though.

Answer 7

It looks disgusting, but if you apply ratio test, I believe things should cancel out nicely. It usually always does :)

Answer 8

Really?

Answer 9

Piece of cake, It converges so quickly you can just try a few terms.

Answer 10

Well.... I would like to do this without a calculator unless the n value is small.

Answer 11

I don't see cake anywhere @tkhunny

Answer 12

No I can't solve this @abb0t

Answer 13

Okay I don't get how to apply the ratio test to this.

Answer 14

if someone hasn't helped you in 5 mins i will guide you step by step. I am going to switch to my computer instead of my tablet.

Answer 15

Allright. Thanks :) .

Answer 16

Ok, so do you know the ratio test?
\[L = \lim_{n \rightarrow ∞} \left| \frac{ a_{n+1} }{ a_n } \right|\]

Answer 17

Yep I know that.

Answer 18

Ok, so you have: \[\left| \frac{ (-1)^{n+1} }{ 9^{n+1}(n+1)! } \times \frac{ 9^nn! }{ (-1)^n } \right| = \left| \frac{ (-1)^n(-1) }{ 9^n9 (n!)(n+1) } \times \frac{ 9^nn! }{ (-1)^n } \right|\]

I'm not going to finish it off for you, but you can see that you can cancel out some terms.

Answer 19

What the... Why did you multiply?

Answer 20

\[\frac{ a }{ \frac{ c }{ d } } = a \times \frac{ d }{ c }\]

Answer 21

SO I wnd up getting -1/(9(n+1))

Answer 22

So I end up*

Answer 23

Anyways, when you're done, the final step is to take the limit and find L.
Once you find L, you should be able to see whethere the series converges or diverges.
Remember:
L < 1 = absolutely diverges
L > 1 = diverges
L = 1 test is inconclusive

Answer 24

Well the limit is 0 which is less than one so the series converges.

Answer 25

But how does that tell me the number of terms?

Answer 26

@abb0t

Answer 27

do as @tkhunny 1st posted. and use a calculator

Answer 28

I got 4 which is incorrect.

Answer 29

Wel... 3.02 or something but we round that to 4.

Answer 30

your n should be equal to 5 so that 1/(9^n*n!) < 5x10^-6
the number of terms should be 6

Answer 31

http://www.wolframalpha.com/input/?i=200000%3D+9%5E%28n%2B1%29%28n%2B1%29%21

Answer 32

if you use your calculator, compute \[\frac{1}{9^n\cdot n!}\]
what n will yield a value less than 5x10^(-6) ?

Answer 33

if n=4, 1/(9^4*4!)=6.35x10-6
if n=5, 1/(9^5*5!) = 1.41x10-7<5x10-6

therefore n = 5
because the lower index begins at zero, (n=0), you will need 6 terms (n=0,1,...,5)

Answer 34

Ohh! Thank you :) .

Answer 35

No it's wrong.

Answer 36

Ohh well. Thanks anyways.