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tiffanymak1996
Group Title
is this absolutely convergent or conditionally or divergent?
 one year ago
 one year ago
tiffanymak1996 Group Title
is this absolutely convergent or conditionally or divergent?
 one year ago
 one year ago

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

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
Divergent...
 one year 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?
 one year 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?
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
T.T
 one year 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?
 one year ago

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

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

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

tiffanymak1996 Group TitleBest ResponseYou've already chosen the best response.0
ok...
 one year 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?
 one year 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?
 one year ago

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

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

godorovg Group TitleBest ResponseYou've already chosen the best response.1
welcome ...
 one year ago

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