Question

Does the series converge?

(-1)^(n-1)/(sqrt(3n-1)) n=1 to infinity

Answer 1

this is an alternating series, all that is needed is that the terms go to zero

Answer 2

how did u know it was an alternating series?

Answer 3

because of the \((-1)^{n-1}\) which is \(1\) if \(n\) is odd and \(-1\) if \(n\) is even

Answer 4

so then do i take the limit of it?

Answer 5

no, you only need to check that the terms go to zero, which they do because, well because they do
you have \(\frac{1}{\sqrt{3n-1}}\)
the numerator is 1, and the denominator has an \(n\) in it, so the terms go to zero

Answer 6

i mean i guess you have to check it, but really you just have to say it:
\[\lim_{n\to \infty}\frac{1}{\sqrt{3n-1}}=0\]

Answer 7

okay awesome:)!!! thank you....now will it be that easy every time?

Answer 8

i can imagine a harder example, but if the series alternates, and the terms go to zero, then the sum converges

Answer 9

for example
\[\sum\frac{1}{\ln(x)}\] certainly does not converge, but
\[\sum \frac{(-1)^n}{\ln(n)}\] does

Answer 10

so do most alternating series converge?

Answer 11

as long as the terms go to zero they do
of course in your last example it would not, because the terms did not go to zero

Answer 12

it is a much easier check that all the other stuff,
integral test
ratio test
root test
etc

Answer 13

yeah the ratio test is easy and the limit test

Answer 14

btw you should recognize for a test or whatever that if you see \((-1)^n\) then the series alternates

Answer 15

okay awesome....the next problem however is telling me that it converges but is asking how many terms give a partial sum within .01 so do i just plug in terms and see which one will cross over .1 to tell me how many sums there are?

Answer 16

hmm now i have to think
that is certainly one way to do it by checking the numbers

Answer 17

i think the error is no more that the absolute value of the last term

Answer 18

what is the actual question? maybe we can do it

Answer 19

Find how many terms give a partial sum, Sn, within 0.01 of the sum, S, of the series.

(-1)^(n-1)/(2n)! n=1 to infinity

Answer 20

i am pretty sure that your job here is to find \(n\) so that \(\frac{1}{(2n!)}<.01\)

Answer 21

you will need that many terms

Answer 22

it won't be large that is for sure

Answer 23

i think \(n=3\) will do it, since \(\frac{1}{6!}=\frac{1}{720}<.01\)

Answer 24

did u just pick 3 randomly?

Answer 25

no i tries \(2\) first but \(\frac{1}{4!}=\frac{1}{24}\) is not small enough

Answer 26

on the other hand we know \((2n)!\) gets real large real fast, so i figured to start small,
i suppose you could say it was a guess

Answer 27

okay !!

Answer 28

if it was \(\sum\frac{(-1)^n}{\sqrt{n}}\) we would have to solve \(\frac{1}{\sqrt{n}}<.01\) so in that case you would need a larger \(n\) like say \(n=1000\)

Answer 29

actually \(n=10,000\)

Answer 30

where did the square root come from?