Question

Need help with this series.
See comment for equation.

Answer 1

http://www.wolframalpha.com/input/?i=Sum%20of%20%5B%2F%2Fmath%3A((-1)^n)(n%2F(n%2B1))%2F%2F%5D%20from%20%5B%2F%2Fmath%3A0%2F%2F%5D%20to%20%5B%2F%2Fmath%3Ainfinity%2F%2F%5D

Answer 2

need to know if it converges/ diverges
and if it converges, what the sum is

Answer 3

also, what test is used to find it

Answer 4

well you can see it's alternating

Answer 5

however it fails the alternating test

Answer 6

it obviously diverges as you can use the nth term

Answer 7

the nth term?

Answer 8

you know, if the limit does not = 0 then it diverges

Answer 9

You can use the nth term test for divergence. Take the limit as n goes to infinity.
\[\lim_{n \rightarrow \infty}\left| \frac{n}{n+1} \right| \implies1\]
Nth term test:
Consider the series:
\[\sum_{n=1}^{\infty}a_n\]
Also consider:
\[\lim_{n \rightarrow \infty} \sum_{n=1}^{\infty}a_n=A\]
The only way a series can converge is if:
\[A \rightarrow 0\]
If it goes to ANY OTHER VALUE or does not exist, then the series diverges.

Answer 10

ok that makes sense now

Answer 11

HOWEVERRRR you cannot use the nth term test to test for convegence as this test does not show convergence

Answer 12

so if it = 0 then this test fails

Answer 13

thats probably one of the few things i know about series

Answer 14

Good :D At least you have some footing :P

Answer 15

i have another one
1/n(n-1)
from 3 to infinity

Answer 16

Do you know the limit comparison test?

Answer 17

yes, but very limited knowledge of it

Answer 18

Okay well you have a telescopic series. Do you know how to do partial fraction decomposition? I mispoke, limit comparison won't work.

Answer 19

misspoke**

Answer 20

wolfram says integral tests works
and it converges to 1/2

Answer 21

just dont jnow how to get to that

Answer 22

But see, if you do the integral test, it doesn't tell you what it converges TO, it just tells you it converges. Treating it as a telescopic series will get you the actual sum. Do you know partial fractions?

Answer 23

yes

Answer 24

Do partial fraction decomp on the series, and tell me what you get :P

Answer 25

i got
-1/n + 1/(n-1)

Answer 26

I get:
\[\sum_{n=3}^{\infty}\frac{1}{n-1}-\frac{1}{n}\]
Nice.

Answer 27

So basically, you want to plug in values of n (starting at 3) until a pattern develops. So lets do that.
Starting at n=3 you have:
\[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+...(\frac{1}{n-2}-\frac{1}{n-1})+(\frac{1}{n-1}+\frac{1}{n})+...\]
Do you see the pattern? The 1/3s,1/4s,1/5s, etc cancel leaving only the first term (1/2) and the LAST term 1/n. So take the limit as n goes to infinity giving:
\[\lim_{n \rightarrow \infty} \frac{1}{2}+\frac{1}{n} \rightarrow \frac{1}{2}\]
You can look up telescopic series to get more examples. I hope that made sense. Also, just a hint, if your teacher puts: "if it converges, find its sum" its either telescopic or geometric :P They are the only ones they generally expect students to do. Finding the nth partial sums (like we just did) is hard and tedious in general. Telescopic series are just nice. If this makes sense to you, I'm going to hop off of here, 10:10 class tomorrow :/

Answer 28

thank you so much :D

Answer 29

No problem :P I hope my explanation helps, I'm alot better at tutoring in person :P Night xDD