Question

Help with a series problem

Answer 1

\[\sum_{n=1}^{n=\infty} \frac{ 3 }{ n(n+3)}\]

Answer 2

how to determine if it converges or diverges?

Answer 3

if im not mistaken (i suck with series) but isnt this an alternating series?

Answer 4

eyeball test
the degree of the denominator is 2
the degree off the numerator is 0
since 2 - 0 = 2 and \(2>1\) then it converges

Answer 5

But how to find out what it converges too? And is it actually called the eyeball test :p

Answer 6

integral test proves this, but it is always true for a rational expression that if the degree of the denominator is greater than the degree of the numerator by more than one (strictly greater than 1) the series converges

Answer 7

interesting

Answer 8

but can u still use the alternating series test?

Answer 9

and no, it is not actually called the eyeball test, but you really can eyeball it that way, no work involved
as for the actual sum that is a different matter
partial fraction decomposition should give it to you
you will find a collapsing sum

Answer 10

no, it does not alternate

Answer 11

I did partial fraction decomp but not sure where to go from there

Answer 12

it does telescope i guess, but that is not the same thing as alternating
write out the first few terms and you will see it
all will be killed off but the first one or two

Answer 13

Ohhhh
Comes to 11/6

Answer 14

I have a bad habit of not writing out the terms

Answer 15

But another question

Answer 16

about this problem

Answer 17

i don't know i didn't do it, but i think it is
\[\frac{1}{n}-\frac{1}{n+3}\] see what survives when it collapses
probably just the first couple

Answer 18

When i decomp it to (1/n)-(1/(n+3)) doesnt 1/n diverge ? and 1/n+3 diverge?

Answer 19

oh no

Answer 20

because i thought 1/n always diverged

Answer 21

you are not summing \(\frac{1}{n}\)

Answer 22

write them out and it will be clear

Answer 23

i.e. start with \(n=1,2,3,4\) it will be obvious that only the first couple of terms survive when you do the subtraction

Answer 24

But why wouldn't you perform the nth term test on it?

Answer 25

because it is not \(\frac{1}{n}\) it is \(\frac{3}{n(n+3)}\)
when you decompose it as
\[\frac{1}{n}-\frac{1}{n+3}\] you are in the form \(\infty-\infty\)
which is undetermined
at the risk of repeating myself, write it out, it will be clear

Answer 26

oh ok ok sorry just needed to be sure, I was a little fuzzy on that part. Thank you

Answer 27

no problem
you did write them out right?

Answer 28

Yes, I did i saw only 1,1/2, and 1/3 was left thanks @satellite73