Question

The series converges to?

Answer 1

\[\Huge \sum_{k=2}^\infty \frac{1}{k(k-1)}\]

Answer 2

First, of course, we should ask ourselves if it Converges at all! Have you yet proven that?

Have you considered a Partial Fraction decomposition?

Answer 3

nope i dunno how to do these type of questions,please guide me

Answer 4

I just did. 1) Does it converge? Prove it!

Answer 5

what does convering mean? has a value?

Answer 6

Converging, it means, after you sum up the sequence, it doesn't get any bigger than a certain number, roughly speaking...

In other words, it's limited... (I've never been the best with words, though XD)

Answer 7

okay so how to solve it

Answer 8

DOES it converge? I believe tkhunny left you that question.
Of course, it DOES, but you'll have to prove it.

Answer 9

how can u say by seeing

Answer 10

There are tests that you can use.... most straightforward here would be what's called the Limit Comparison Test, though if you're clever enough, you could use a Direct Comparison...

Answer 11

This is just plain WRONG! How can you be asked to find a sum if you don't know what "convergence" means?! I blame your teachers or course design, not you. It's just utterly and foolishly out of order. Please complain to whoever wrote the curriculum.

Have you EVER met the concept of "Partial Fractions"? After the "Convergence" fiasco, I'm not expecting the answer to be "yes".

New Math Teaching Methods: Just trow the students in a pool of sharks and yell, "Good Luck"!

\(\dfrac{1}{k(k-1)} = \dfrac{1}{k-1} - \dfrac{1}{k}\)

Answer 12

@INeedHelpPlease?
HAVE you taken calculus?

Answer 13

i am in high school

Answer 14

Well then, at least the concept of partial fractions should not be alien to you, you would have done that as a technique of integration, no?

Answer 15

yep i know that method,what next?

Answer 16

Is it convergent, though?
Okay, since you don't seem familiar with the tests (correct me if I'm wrong)
Then we'll just have to do this with brute force... for now.

Answer 17

Let
\[\Huge s_n=\sum_{k=2}^n\frac{1}{k(k-1)}\]

Answer 18

Do you understand this notation?

Answer 19

yes

Answer 20

Okay, so that we have
\[\Huge \lim_{n\rightarrow \infty}s_n=\sum_{k=2}^\infty \frac{1}{k(k-1)}\]

Answer 21

Now, I want you to find \(\large s_2\)

Answer 22

hmm

Answer 23

i just somewhat understand the notation :|

Answer 24

That's a problem.

Answer 25

yep it is :P

Answer 26

It's not one to be taken lightly, though... are you sure you're ready for this kind of question? D:

Answer 27

but i have done all this beofre,long back,just dont remember maybe i get it after seeing a bit

Answer 28

Or reviewing a bit. Go ahead, take your time and review...