Question

determine the convergence or divergence of the sequence with the given nth term. If the sequence converges, find its sum. a(sub)n= (n-2)!/n!

Answer 1

start with
\[\frac{(n-2)!}{n!}=\frac{1}{n(n-1)}\] and you should be in good shape

Answer 2

i did that and then did partial fractions but i don't know how the sum equals one

Answer 3

first off i hope we are starting at 2 because it is not defined if you start at n = 1

Answer 4

it doesn't say in the problem

Answer 5

but one and zero would both make a zero in the denominator so we'd have to start at two

Answer 6

Err, nevermind that. :p

Answer 7

sorry that was a stupid question i asked, since the numerator is (n-2)!

Answer 8

haha it's telescoping, i just didn't see that before. thank you!

Answer 9

yes it sure is! you can see that from the partial fractions right?

Answer 10

yeah, i just missed that before because i didn't do enough of the sums out. how do we know that it converges though?

Answer 11

that pretty much tells you right there. of course could also compare to
\[\sum\frac{1}{n^2}\] which you know converges

Answer 12

or just say what the partial sums are

Answer 13

yeah because p is greater than one.
can you help me with \[\sum_{n=2}^{\infty} 1/n \sqrt{n ^{2}-1}\]

Answer 14

you mean find the sum?

Answer 15

determine if it converges or diverges and state the test used

Answer 16

it is a set up for the integral test i believe

Answer 17

i thought so but then i wasn't sure how to take the integral

Answer 18

\[u=\sqrt{x^2-1}, du=\frac{x}{2\sqrt{x^2-1}}\]

Answer 19

ok that was wrong, sorry

Answer 20

for some reason i can't see the equation before i post

Answer 21

oh that's weird

Answer 22

but that is the idea, use a u - sub with
\[u=\sqrt{x^2-1}\]

Answer 23

but if you have the square root in the u- sub then you have to do chain rule

Answer 24

yeah i messed up

Answer 25

damn typos

Answer 26

haha it's fine, you're being a huge help! i have a quiz tomorrow..

Answer 27

good luck

Answer 28

wait lets try this, since i can't type

Answer 29

here, click on show steps and it will give you the integral. u sub for sure

Answer 30

http://www.wolframalpha.com/input/?i=+integral+1%2F%28xsqrt%28x^2-1%29%29

Answer 31

crap i still haven't memorized those arctans etc

Answer 32

do you mind if i ask you another?

Answer 33

hmm maybe another method but it would amount to a trig sub and you would get arctan in any case

Answer 34

you can ask, but it might be best just to post an new question. you will get lots of responses

Answer 35

i only need one right response haha

Answer 36

Find the sum of the convergent series \[\sum_{n=1}^{\infty} 1/n(n+3)\]