Question

Zarkon, more series help needed :/

Answer 1

bahrom, you keep asking for serious help :P

Answer 2

Sorry to bother u guys, but i've been working on series non stop for the whole day today, and i can't really solve theoretical stuff right now - only computational. How would I do this question?

Answer 3

@Zarkon another one!

Answer 4

\[a_n=\frac{1}{n}\]

\[\sum_{n=1}^{\infty}\left[\frac{1}{n+1}-\frac{1}{n}\right]\]

\[s_n=\frac{1}{2}-\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\cdots...\frac{1}{n+1}-\frac{1}{n}\]

\[s_n=-1+\frac{1}{n+1}\]
\[s_n\to -1\text{ as }n\to\infty\]

but \[\sum_{n=1}^{\infty}\frac{1}{n}=\infty\]

Answer 5

ty guys!

Answer 6

bahrom, what course are you taking?

Answer 7

this is not mine, im helping a friend. This is calc 2.. I forgot so much stuff lol, and that was only 2 yrs ago.

Answer 8

hahahaha

Answer 9

one more series incoming zarkon

Answer 10

Pippa, is something funny?

Answer 11

Yes lol

Answer 12

Would you mind sharing the joke with the rest of us? I didn't get it.

Answer 13

haha most things that i think r funny nobody else does

Answer 14

You didn't answer my question.

Answer 15

umm meaning i dont want to share it

Answer 16

Well, you should have laughed to yourself then. Why share the laugh and not the joke?

Answer 17

LOL

Answer 18

hmmm sorry abt that

Answer 19

haha bahrom laughed at that

Answer 20

@samjordon what a coincidence that we are viewing this at the same time

Answer 21

3 people on such an old question? o.O

Answer 22

who is 3?

Answer 23

Ishaan was....he left

Answer 24

ohhhh lol this was one of my old questions so i recognized it

Answer 25

Lol why are you on this question btw?

Answer 26

hahahah cuz i saw u on these zarkon last question and all these series questions. Like it appeared next to ur name and i recognized them as mine so i followed u here