Question

Determine convergence.

Answer 1

\[\sum_{n=1}^{\infty}\frac{ -1 }{ n+3 }+\frac{ 1 }{ n+1 }\] should I just take the limit?

Answer 2

@jim_thompson5910 :)

Answer 3

It appears to go to 0 which means it diverges, but i'm not sure how to show that algebraically

Answer 4

put the plus first, minus second

Answer 5

\[\sum \frac{1}{n+1}-\frac{1}{n+3}\] start adding

Answer 6

Start adding what?

Answer 7

put \(n=1\) then \(n=2\) then \(n=3\) etc

Answer 8

it will be clear what you get when you do it

Answer 9

a_1=1/4
a_2=2/15
a_3=1/12
a_10=0.13
a_20=0.004

Answer 10

How many should I do? xD

Answer 11

who hold the phone

Answer 12

don't compute anything, just add

Answer 13

i.e. for example if \(n=1\) you get \[\frac{1}{2}-\frac{1}{4}\]

Answer 14

next one is \(n=2\) get \[\frac{1}{3}-\frac{1}{5}\] what is the next one?

Answer 15

1/4-1/6

Answer 16

and the one after?

Answer 17

1/5-1/7

Answer 18

1/6-1/8
n=5

Answer 19

ok now what do we have all together ?

Answer 20

1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7...

Answer 21

a bunch of things add up to zero

Answer 22

151/168 ? lol

Answer 23

\[\frac{ 1 }{ 2 }+\frac{ 1 }{ 3 }-\frac{ 1 }{ 7 }-\frac{ 1 }{ 8 }\] is what we're left with

Answer 24

ok i am done torturing you
all the terms add up to zero, since you have plus and minus for each one

Answer 25

the only thing that does not have a match are the first two positive numbers \[\frac{1}{2}+\frac{1}{3}\]

Answer 26

xD so if we continued the pattern would continue that everything is cancelled out.

Answer 27

i meant those terms do not get cancelled with later ones

Answer 28

so everything else adds to zero, but you still have \[\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\] as your sum

Answer 29

lets see how we would have known this from the start ok?

Answer 30

\[\frac{1}{n+1}\] and \[-\frac{1}{n+3}\] the second one is two steps away from the first
so only the first two non zero terms remain when you add

Answer 31

aaah
Okay. So..does this thing go to 0 or not? I guess that's the only part i'm confused about.
Because the 1/2 and 1/3 will always stay there.

Answer 32

the sum is not zero, the sum is \(\frac{5}{6}\)
that is what you get when you add all that stuff up

Answer 33

Wat. The limit goes to 0

Answer 34

the limit of what ?

Answer 35

i think you are getting the series (what you get when you add) the the sequence of the terms

Answer 36

http://www.wolframalpha.com/input/?i=lim+n+approaches+infty+((1%2F(n%2B1))-(1%2F(n%2B3))

Answer 37

of course the terms go to zero
they have to if the sum is to be finite
in fact that is not enough for the sum to exist

Answer 38

But you just said this goes to 1/2+1/3
which would make it divergent.
But if it goes to 0 that makes it convergent ..

Answer 39

http://www.wolframalpha.com/input/?i=sum+((1%2F(n%2B1))-(1%2F(n%2B3))+n%3D1+to+infinity

Answer 40

nvm I got it. Thanks man!