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tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} (1)^{n}\times e ^{\frac{ 1 }{ n }}\]

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0yes that's the answer given, but i got conditionally throughout the process can u go thru the steps with me?

Herp_Derp
 2 years ago
Best ResponseYou've already chosen the best response.0Well, 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?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0but, i thought n=1 under the sigma sign...

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1No, tiffany they are telling right here, it is diverges

jeanlouie
 2 years ago
Best ResponseYou've already chosen the best response.0we 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

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, what i meant was i know it is divergent, but when i go thru the steps i got conditionally convergence

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0don't understand this part an = (−1)n×e1/n , then  a2n  a2n+1  > 1 with every n > some N it is diverges

jeanlouie
 2 years ago
Best ResponseYou've already chosen the best response.0i mean, for example, let N = 1000000, then for every n > N, a_2n  a_(2n+1) > 1

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1Jeanlouie 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?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0for the ratio test if the lim is <1 then the series converges, if lim >1 or=infinity the series diverges, if lim=1 inconclusive

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0i 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.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0but 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.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0although i know i'm wrong, cause the answer is divergent:(

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1what is the value {1} of this when testing ratio =

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0when testing ratio without the absolute value i got 1

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1How does the value change to 1 I am lost there?

jeanlouie
 2 years ago
Best ResponseYou've already chosen the best response.01 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

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0\[\rho = \lim_{n \rightarrow +\infty} \frac{ (1)^{n+1} \times e ^{\frac{ 1 }{ n }}}{ ((1)^{n})\times e ^{\frac{ 1 }{ n }}\ }\]

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0ln both top and bottom

jeanlouie
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, ρ = 1, you got it

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0becomes:\[\lim_{n \rightarrow +\infty} \frac{ \frac{ 1 }{ n+1} }{ \frac{ 1 }{ n } } \]

jeanlouie
 2 years ago
Best ResponseYou've already chosen the best response.0no, top or bottom may be negative, we can't use ln here.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0(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

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1Tiffany I think you have the right answer.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0i do? but this is like conditionally ; yet the modal answer is divergent...

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1Tiffany 1+1n is not 1 it is 2n

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1you see where this is?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0in 1/(1+1/n), if i substitute n as infinity, then 1/infinity is 0, so 1/(1+0)=1/1

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0is this correct?

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1when sub in n, you still have 1+1 which is 2.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0where does the 1+1 com from?

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1so 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?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0so sorry, it should be 1/(1+(1/n)) not 1/(1+1/n)

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow +\infty}\frac{ 1 }{ 1+\frac{ 1 }{ n } }\]

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0so... \[\lim_{n \rightarrow +\infty} \frac{ 1 }{ 1+\frac{ 1 }{ n } }\]

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1Tiffany can the bottom be 0? or am I missing something here?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0the bottom can't be zero, what i did is just divide top and bottom with n

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0and then: \[\frac{ 1 }{ 1+\frac{ 1 }{ \infty } }\]

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0and since \[\frac{ 1 }{ \infty }=0\],

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 1+0 }=1\]

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0and i did this like 5 times and still don't get why it's divergent...

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1what happens if we 1/1+0=1 now what if simply the bottom to 1/1=1 than what?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0if it's 1 then the series is convergent, not divergent.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0well, the model answer IS divergent.

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1only if stays at one, 1=1 if you add one you get 0

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0and that's what frustrates me the most.

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1so I am taking the answer can't be 0 for the final answer?

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0no, it's not that it can't be 0, just that how do u suddenly +1 after the limit?

godorovg
 2 years ago
Best ResponseYou've already chosen the best response.1Because the +1 falls within the limit.

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0oh, ok, thx, i think i get it now. :)

tiffanymak1996
 2 years ago
Best ResponseYou've already chosen the best response.0sry for bothering you so long...
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