Question

is this absolutely convergent or conditionally or divergent?

Answer 1

\[\sum_{n=1}^{\infty} (-1)^{n}\times e ^{\frac{ 1 }{ n }}\]

Answer 2

Divergent...

Answer 3

yes that's the answer given, but i got conditionally throughout the process can u go thru the steps with me?

Answer 4

Well, for some large \(n\), \(e^{1/n}\) is sufficiently close to \(1\) so that we can compare this series to the divergent series:\[\sum_{n=0}^{\infty}(-1)^n\]The larger the \(n\), the closer your series is to this series. Therefore your series diverges. Do you understand why \(\sum_{n=0}^{\infty}(-1)^n\) diverges?

Answer 5

but, i thought n=1 under the sigma sign...

Answer 6

No, tiffany they are telling right here, it is diverges

Answer 7

we have
abs(\[(-1)^{n}\times e ^{1/n}\]) -> 1 when \[n \rightarrow \infty\]
but if consider \[a_{n}\] = \[(-1)^{n}\times e ^{1/n}\] , then
|\[a_{2n}\] - \[a_{2n+1}\]| > 1 with every n > some N
it is diverges

Answer 8

yeah, what i meant was i know it is divergent, but when i go thru the steps i got conditionally convergence

Answer 9

don't understand this part
an
=
(−1)n×e1/n
, then
|
a2n
-
a2n+1
| > 1 with every n > some N
it is diverges

Answer 10

i mean, for example, let N = 1000000, then for every n > N, |a_2n - a_(2n+1)| > 1

Answer 11

Jeanlouie I haven't this type calc for a while since High School, in order for this to be diverges it means the n is closer to 1 or I am lost here?

Answer 12

for the ratio test if the lim is <1 then the series converges, if lim >1 or=infinity the series diverges, if lim=1 inconclusive

Answer 13

i know that testing the series for absolutely convergent or conditionally or divergent is to try the ratio test with absolute value first then the one without.

Answer 14

but when i calculate the one with absolute value the limit =1 so inconclusive; when i test the series for the ratio test without absolute value the limit =-1, so it's convergent; that's how i get conditionally convergent.

Answer 15

although i know i'm wrong, cause the answer is divergent:(

Answer 16

what is the value {1} of this when testing ratio =

Answer 17

when testing ratio without the absolute value i got -1

Answer 18

How does the value change to -1 I am lost there?

Answer 19

1 is only an example here, 0.5 or any other \[\epsilon < -2\] is OK, too.
when \[n \rightarrow \infty \], \[a_{2n} \rightarrow 1\] and \[a_{2n+1}\rightarrow -1\]


ps:sr for my poor English

Answer 20

ϵ< 2, sorry

Answer 21

\[\rho = \lim_{n \rightarrow +\infty} \frac{ (-1)^{n+1} \times e ^{\frac{ 1 }{ n }}}{ ((-1)^{n})\times e ^{\frac{ 1 }{ n }}\ }\]

Answer 22

ln both top and bottom

Answer 23

yeah, ρ = -1, you got it

Answer 24

becomes:\[\lim_{n \rightarrow +\infty} -\frac{ \frac{ 1 }{ n+1} }{ \frac{ 1 }{ n } } \]

Answer 25

no, top or bottom may be negative, we can't use ln here.

Answer 26

(the limit is still there but i'm just lazy:
then:
-n/(n+1)
divide top and bottom with n:
-1/(1+1/n)

substitute infinity:
-1/(1+0)
=-1

Answer 27

Tiffany I think you have the right answer.

Answer 28

i do? but this is like conditionally ; yet the modal answer is divergent...

Answer 29

Tiffany 1+1n is not -1 it is 2n

Answer 30

model, sry

Answer 31

2/n not -1

Answer 32

you see where this is?

Answer 33

in -1/(1+1/n), if i substitute n as infinity, then 1/infinity is 0, so -1/(1+0)=-1/1

Answer 34

is this correct?

Answer 35

when sub in n, you still have 1+1 which is 2.

Answer 36

where does the 1+1 com from?

Answer 37

so if we look at this -1(1+1/n) would give you add the two ones which would 2/n times that by -1 and what is the answer?

Answer 38

so sorry, it should be -1/(1+(1/n)) not -1/(1+1/n)

Answer 39

-1/1=-1times +1 =-1/n

Answer 40

\[\lim_{n \rightarrow +\infty}\frac{ -1 }{ 1+\frac{ 1 }{ n } }\]

Answer 41

yes..

Answer 42

so...
\[\lim_{n \rightarrow +\infty} \frac{ -1 }{ 1+\frac{ 1 }{ n } }\]

Answer 43

Tiffany can the bottom be 0? or am I missing something here?

Answer 44

the bottom can't be zero, what i did is just divide top and bottom with n

Answer 45

and then:
\[\frac{ -1 }{ 1+\frac{ 1 }{ \infty } }\]

Answer 46

and since \[\frac{ 1 }{ \infty }=0\],

Answer 47

\[\frac{ -1 }{ 1+0 }=-1\]

Answer 48

and i did this like 5 times and still don't get why it's divergent...

Answer 49

T.T

Answer 50

what happens if we -1/1+0=-1 now what if simply the bottom to -1/1=-1 than what?

Answer 51

if it's -1 then the series is convergent, not divergent.

Answer 52

yes..

Answer 53

well, the model answer IS divergent.

Answer 54

only if stays at one, -1=-1 if you add one you get 0

Answer 55

and that's what frustrates me the most.

Answer 56

ok...

Answer 57

so I am taking the answer can't be 0 for the final answer?

Answer 58

no, it's not that it can't be 0, just that how do u suddenly +1 after the limit?

Answer 59

Because the +1 falls within the limit.

Answer 60

oh, ok, thx, i think i get it now. :)

Answer 61

welcome ...

Answer 62

sry for bothering you so long...