Question

How can I show the series from 0 to infinity (1/n+3)-(1/n+1) converges?

Answer 1

\[\sum_{n=0}^{\infty}\frac{ 1 }{ n+3}-\frac{ 1 }{ n+1}\]

Answer 2

Have you tried the comparison test

Answer 3

I have. But I'm confused because the series is always negative. I'm not sure what I could compare it to.

Answer 4

You should consider writing out a few terms of the series and see if you notice any patterns.

Answer 5

I'll let kai handle this one xD

Answer 6

Nope, that's all I'm saying lol I'm busy atm

Answer 7

hmmmm so its getting increasingly less negative? Looks like it converges to 0 maybe?

Answer 8

But I'm still not sure what I could compare it to...

Answer 9

It's a telescoping series, just do what kai said and you should get it ;d

Answer 10

Maybe 1/n^2

Answer 11

I thought about that but 1/n^2 is always larger than this function.

Answer 12

You don't compare it to anything since you can exactly evaluate it.

Answer 13

Oh wait, I said that wrong.

Answer 14

http://www.wolframalpha.com/input/?i=series+from+0+to+infinity+1%2F%28n%2B3%29-1%2F%28n%2B1%29

Answer 15

Exactly evaluate it? how?

Answer 16

try writing out the first few terms as @Kainui suggested

Answer 17

^

Answer 18

you should see a lot of zeroing out

Answer 19

For the first three terms I got -.66, -.25, and -.133

Answer 20

i would leave them infractional form

Answer 21

in fractional *

Answer 22

I see its getting closer to zero....But thats not enough right?

Answer 23

Oh ok. I'll do that.

Answer 24

-2/3, -1/4,-2/15...

Answer 25

I'm sorry, im still not sure how that helps me =/

Answer 26

well that doesn't help

Answer 27

it looks like you combined the fractions

Answer 28

\[\LARGE \sum_{n=0}^\infty \frac{1}{n+3}-\sum_{n=0}^\infty \frac{1}{n+1}\]

We can see that by replacing n with k+2 that it looks similar for the second summation

\[\LARGE \sum_{n=0}^\infty \frac{1}{n+3}-\sum_{k+2=0}^\infty \frac{1}{k+2+1}\]\[\LARGE \sum_{n=0}^\infty \frac{1}{n+3}-\sum_{k=-2}^\infty \frac{1}{k+3}\] Now we pull out the -2 and -1 terms leaving us at k=0 so that it matches the other one: \[\LARGE \sum_{n=0}^\infty \frac{1}{n+3}-\frac{1}{1}-\frac{1}{2}-\sum_{k=0}^\infty \frac{1}{k+3}\]
but since k and n are just dummy variables, these both represent the exact same summation. So they cancel out leaving us with just a couple terms. Fancy yeah?

Answer 29

ohhhh I'm getting that!!

Answer 30

Here an example:
\[\sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1}) \\ (\frac{1}{1}-\frac{1}{1+1})+(\frac{1}{2}-\frac{1}{1+2})+(\frac{1}{3}-\frac{1}{1+3})+(\frac{1}{4}-\frac{1}{4+1}) +\cdots \\ (\frac{1}{1}-\cancel{\frac{1}{1+1}})+(\cancel{\frac{1}{2}}-\cancel{\frac{1}{1+2}})+(\cancel{\frac{1}{3}}-\cancel{\frac{1}{1+3}})+(\cancel{\frac{1}{4}}-\frac{1}{4+1}) +\cdots \\ =\frac{1}{1}=1 \]
I didn't put a little cancelly thing over the 1/5 part but it does cancel

Answer 31

Ok I think ive got it. Thanks to all that helped out!!

Answer 32

I like the what @Kainui wrote it

Answer 33

way*

Answer 34

Yeah me to... Never would have thought to do it that way.

Answer 35

Yeah, it's a lot like "u-substitution" when solving an integral. Once you know about this way, if you can transform the sum into the other sum by a simple little substitution like this you can sometimes get them to work out pretty nicely like this. =D