Question

Does this sequence diverge or converge?


\[\Large \frac{e^n-e^{-n}}{e^{2n}-1}\]

Answer 1

Converge

Answer 2

xD

Answer 3

sum of that right for n from 0 to inf?

Answer 4

Not sure let me calculate it xD

Answer 5

converg

Answer 6

This isn't series btw

Answer 7

oh..

Answer 8

Then what do you want to know about this

Answer 9

converg

Answer 10

The numerator is 2Sinh x, hyperbolic sin fuction

Answer 11

yeah how did you know dan

Answer 12

seen it before

Answer 13

dan invented math, that's how?

Answer 14

cool

Answer 15

:D

Answer 16

So it converges to 0?

Answer 17

luigi, the top is a function of x and the bottom a function of n, so I am not quite sure what you are asking

Answer 18

are the values of n increasing in a series, otherwise for any values of n and x, you will find certain output value.

Answer 19

Oh crap, I didn't even notice.. they were all suppose to be n..

Answer 20

For the first "series" from 0 to infinity, if you let 2n = 2x so there's only one variable, it equals e/(e-1)

Answer 21

so it converges at 1.58 blah blah blah

Answer 22

okay and its a sum of n from 0 to inf right?

Answer 23

yes, the sum converges to that number

Answer 24

Lets do convergence tests on it to see if it converges still

Answer 25

@Kainui what test?? integral tests?

Answer 26

What? I just got here, I'm not even sure I know what's going on to be honest? Converging how? Like as n approaches 0 or n approaches infinity?

Answer 27

Actually, you don't need a test for this.. just

\[\Large \lim_{x \rightarrow \infty} \frac{e^n-e^{-n}}{e^{2n}-1}\]

Answer 28

Oh i thought it was a series

Answer 29

It does say "sequence" at the question xD

But how would I solve that limit? .-.

Answer 30

okay well you know it will be e^inf/e^(2inf) so its pretty obvious it must converge to 0

Answer 31

Are you allowed to use L'Hopital's rule?

Answer 32

we can divide top and bottom by e^n to see more clearly