Question

Need help with convergence test.

Answer 1

\[\sum_{1}^{\infty} (7^n)/(8^n+3)\]

Answer 2

I thought I would use the series ratio test, and I ended up with
\[7(8^n+3)/8^{n+1} + 3\]
Not sure how I would do this after.

Answer 3

as lim n --> infinity woops.

Answer 4

so you tried ratio test?
hmmm...
\[\lim_{n \rightarrow \infty}|\frac{7^{n+1}}{8^{n+1}+3} \cdot \frac{8^n+3}{7^n}|=\lim_{n \rightarrow \infty}|\frac{7(8^n+3)}{8^{n+1}+3}|=\lim_{n \rightarrow \infty}|\frac{7(8^n)}{8^{n+1}}|\]

Answer 5

do you know why that last step 8^n+3 is basically 8^n for very large n

Answer 6

8^(n+1)+3 is also 8^(n+1) for very large n

Answer 7

does that help you?

Answer 8

\[7(8^n)\div (8n)(8)\]

Answer 9

i mean 8^n

Answer 10

and the 2 8^n's cancel out?

Answer 11

which results in it being 7/8 which is less than 1 which means its convergent

Answer 12

sounds awesome :)

Answer 13

you know if you didn't want to do what I did
you know saying stuff like a+3 is approximately a
for very large a

you could have done l'hospital too

Answer 14

it is just a tad more work

Answer 15

I'm just trying to prep myself up for my test coming up. It's just hard knowing when to use which test and thinking out of the box like that...

Answer 16

\[\lim_{n \rightarrow \infty} \frac{7(8^n+3)}{8^{n+1}+3}=\lim_{n \rightarrow \infty}\frac{7(\ln(8) 8^n+0)}{8\ln(8)8^n+0}=\lim_{n \rightarrow \infty}\frac{7 \ln(8)8^n}{8 \ln(8)8^n}=\frac{7}{8}\]

Answer 17

by l'hosptial

Answer 18

if you prefer that way

Answer 19

Do you know how I would be able to do \[\sum_{2}^{\infty} 3/(n(\ln(n))^2)\]

Answer 20

\[\sum_{n=2}^{\infty} \frac{3}{n (\ln(n))^2}\]
this is right?

Answer 21

do you know derivative of ln(x)?

Answer 22

Reason I ask is because this is looks like something easy to integrate

Answer 23

\[3 \int\limits_{2}^{\infty} \frac{1}{x} \cdot \frac{1}{(\ln(x))^2} dx\]

Answer 24

you have talked about integral test correct?

Answer 25

I see what you did. So this would be the integral test.

Answer 26

yep I know what you mean

Answer 27

its just that I don't know which test to use for each different problem...

Answer 28

well for that first problem that doesn't look easy to integrate
so I would think ratio or root for the first one
let me try to root test on the first one
\[\lim_{n \rightarrow \infty}|\frac{7^n}{8^n+3}|^\frac{1}{n}=\lim_{n \rightarrow \infty}|\frac{7^n}{8^n}|^\frac{1}{n}=\lim_{n \rightarrow \infty}((\frac{7}{8})^n)^\frac{1}{n}=\lim_{n \rightarrow \infty}(\frac{7}{8})^\frac{n}{n}=(\frac{7}{8})^1=\frac{7}{8}\]
it takes practice and sometimes it is trial and error to figure out which test works best
http://tutorial.math.lamar.edu/Classes/CalcII/RootTest.aspx
look at this site
paul has awesome examples

Answer 29

sometimes it is easy to use more than one test as we shown above for the first problem you wanted to do

Answer 30

alright thanks a lot! I have to practice more...

Answer 31

can I message you if I need anymore help later?

Answer 32

ok

Answer 33

If I'm around I will definitely come running.
You can mention me in your question or message me a link to your question. Either way.

Answer 34

@freckles
Hey this is more about sequences
\[A _{n} = \frac{ nsin(n) }{ n^2+1 } \]
Sin can only be between -1 and 1... That means it will be infinity * k which is between -1 to 1.
The bottom will be infinity ^2 which is larger than the numerator meaning the limit will be 0 , which means it will be convergent?

Answer 35

\[-1 \le \sin(n) \le 1 \\ -n \le n \sin(n) \le n \\ \frac{-n}{n^2+1} \le \frac{n \sin(n)}{n^2+1} \le \frac{ n }{n^2+1} \\ \lim_{n \rightarrow \infty}\frac{n}{n^2+1}=0 \text{ and } \lim_{n \rightarrow \infty}\frac{-n}{n^2+1}=0 \\ \text{ so yes } \lim_{n \rightarrow \infty} \frac{ n \sin(n)}{n^2+1}=0\]

Answer 36

that is by squeeze theorem

Answer 37

Thanks ! I'll try writing it how you explained it on the test if it shows up