Question

determine convergence of divergence:

(2/5)+(2*6)/(5*8)+(2*6*10)/(5*8*10)+(2*6*10*14)/(5*8*11*14)+...

Answer 1

Just looking at the pattern, Im hoping there was a slight typo

Answer 2

damn i wrote it wrong....the third term on the demoniator should be (2*6*10)/(5*8*11)

Answer 3

Bleh, just need a way to turn this into a series in terms of n.

Answer 4

well since the first term is (2/5) can we factor that out some how? ANd it also looks like a factorial with something being added....Idk

Answer 5

Yeah, Ive been messing with factorials, too. As far as the 2/5, it just makes me think that it's left over due to n = 0 making things become 1. x^0 or (0)! could cause that. So trying to get ideas of how to mess with that.

Answer 6

Makin my head hurt, haha. It just looks like introduction to power series, but brain isnt coming up with anything.

Answer 7

I'll stick around to see if someone can point us in the right direction, but brain needs to relax xD

Answer 8

totally undertand...thanks for your help anyways !!!

Answer 9

Your series diverges.
Each term is the previous term times a fraction. After the fourth term these multulpliers become
\[
\frac {14} {14}, \ \frac {18}{17}, \frac {22}{20}, \frac {26}{23}, \frac {30}{26}
\]
The sequence of n-terms become increasing, so it cannot go to zero. So the series is divergent

Answer 10

\[
a_1=\frac 2 5\\
a_2 = a_1\frac 6 8\\
a_3 = a_2\frac {10}{ 11}\\
a_4 = a_3\frac {14}{ 14}=a_3\\
a_5 = a_4\frac {18}{ 17}> a_4\\
a_6 = a_5\frac {22}{ 20}>a_5\\

\cdots
\]

Answer 11

@Psymon

Answer 12

Elias thank you very much for your help!!!

Answer 13

Is there a test that you can use to prove the divergence, though? I can see that it will diverge by regular analysis, but I would think that you'd have to say you used some test, right?

Answer 14

Yes, this is called the divergence test.
If \( lim_{n->\infty} a_n \ne 0\) then \[
\sum_1^{\infty} a_n
\]
is divergent.

Answer 15

And what you were showing was proof enough of that? I mean, I see it proves it, but Id be paranoid about whether or not it was considered rigorous enough I suppose.

Answer 16

Believe me, it is an excellent proof of it.
I showed that \( lim_{n\to \infty} a_n \ne 0 \) and that is enough to show that the series is not convergent. See
http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTests/divergence.html