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tiffanymak1996
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is this absolutely convergent or conditionally or divergent?
 2 years ago
 2 years ago
tiffanymak1996 Group Title
is this absolutely convergent or conditionally or divergent?
 2 years ago
 2 years ago

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

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
Divergent...
 2 years ago

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

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

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

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

jeanlouie Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

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

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

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

godorovg Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

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

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

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

jeanlouie Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

jeanlouie Group TitleBest ResponseYou've already chosen the best response.0
ϵ< 2, sorry
 2 years ago

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

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

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

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

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

tiffanymak1996 Group TitleBest 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
 2 years ago

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

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

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

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
model, sry
 2 years ago

godorovg Group TitleBest ResponseYou've already chosen the best response.1
2/n not 1
 2 years ago

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

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

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

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

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

godorovg Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

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

godorovg Group TitleBest ResponseYou've already chosen the best response.1
1/1=1times +1 =1/n
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
ok...
 2 years ago

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

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

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

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

godorovg Group TitleBest ResponseYou've already chosen the best response.1
welcome ...
 2 years ago

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