tiffanymak1996
  • tiffanymak1996
is this absolutely convergent or conditionally or divergent?
Mathematics
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katieb
  • katieb
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tiffanymak1996
  • tiffanymak1996
\[\sum_{n=1}^{\infty} (-1)^{n}\times e ^{\frac{ 1 }{ n }}\]
anonymous
  • anonymous
Divergent...
tiffanymak1996
  • tiffanymak1996
yes that's the answer given, but i got conditionally throughout the process can u go thru the steps with me?

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anonymous
  • anonymous
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?
tiffanymak1996
  • tiffanymak1996
but, i thought n=1 under the sigma sign...
anonymous
  • anonymous
No, tiffany they are telling right here, it is diverges
anonymous
  • anonymous
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
tiffanymak1996
  • tiffanymak1996
yeah, what i meant was i know it is divergent, but when i go thru the steps i got conditionally convergence
tiffanymak1996
  • tiffanymak1996
don't understand this part an = (−1)n×e1/n , then | a2n - a2n+1 | > 1 with every n > some N it is diverges
anonymous
  • anonymous
i mean, for example, let N = 1000000, then for every n > N, |a_2n - a_(2n+1)| > 1
anonymous
  • anonymous
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?
tiffanymak1996
  • tiffanymak1996
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
tiffanymak1996
  • tiffanymak1996
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.
tiffanymak1996
  • tiffanymak1996
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.
tiffanymak1996
  • tiffanymak1996
although i know i'm wrong, cause the answer is divergent:(
anonymous
  • anonymous
what is the value {1} of this when testing ratio =
tiffanymak1996
  • tiffanymak1996
when testing ratio without the absolute value i got -1
anonymous
  • anonymous
How does the value change to -1 I am lost there?
anonymous
  • anonymous
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
anonymous
  • anonymous
ϵ< 2, sorry
tiffanymak1996
  • tiffanymak1996
\[\rho = \lim_{n \rightarrow +\infty} \frac{ (-1)^{n+1} \times e ^{\frac{ 1 }{ n }}}{ ((-1)^{n})\times e ^{\frac{ 1 }{ n }}\ }\]
tiffanymak1996
  • tiffanymak1996
ln both top and bottom
anonymous
  • anonymous
yeah, ρ = -1, you got it
tiffanymak1996
  • tiffanymak1996
becomes:\[\lim_{n \rightarrow +\infty} -\frac{ \frac{ 1 }{ n+1} }{ \frac{ 1 }{ n } } \]
anonymous
  • anonymous
no, top or bottom may be negative, we can't use ln here.
tiffanymak1996
  • tiffanymak1996
(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
anonymous
  • anonymous
Tiffany I think you have the right answer.
tiffanymak1996
  • tiffanymak1996
i do? but this is like conditionally ; yet the modal answer is divergent...
anonymous
  • anonymous
Tiffany 1+1n is not -1 it is 2n
tiffanymak1996
  • tiffanymak1996
model, sry
anonymous
  • anonymous
2/n not -1
anonymous
  • anonymous
you see where this is?
tiffanymak1996
  • tiffanymak1996
in -1/(1+1/n), if i substitute n as infinity, then 1/infinity is 0, so -1/(1+0)=-1/1
tiffanymak1996
  • tiffanymak1996
is this correct?
anonymous
  • anonymous
when sub in n, you still have 1+1 which is 2.
tiffanymak1996
  • tiffanymak1996
where does the 1+1 com from?
anonymous
  • anonymous
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?
tiffanymak1996
  • tiffanymak1996
so sorry, it should be -1/(1+(1/n)) not -1/(1+1/n)
anonymous
  • anonymous
-1/1=-1times +1 =-1/n
tiffanymak1996
  • tiffanymak1996
\[\lim_{n \rightarrow +\infty}\frac{ -1 }{ 1+\frac{ 1 }{ n } }\]
anonymous
  • anonymous
yes..
tiffanymak1996
  • tiffanymak1996
so... \[\lim_{n \rightarrow +\infty} \frac{ -1 }{ 1+\frac{ 1 }{ n } }\]
anonymous
  • anonymous
Tiffany can the bottom be 0? or am I missing something here?
tiffanymak1996
  • tiffanymak1996
the bottom can't be zero, what i did is just divide top and bottom with n
tiffanymak1996
  • tiffanymak1996
and then: \[\frac{ -1 }{ 1+\frac{ 1 }{ \infty } }\]
tiffanymak1996
  • tiffanymak1996
and since \[\frac{ 1 }{ \infty }=0\],
tiffanymak1996
  • tiffanymak1996
\[\frac{ -1 }{ 1+0 }=-1\]
tiffanymak1996
  • tiffanymak1996
and i did this like 5 times and still don't get why it's divergent...
tiffanymak1996
  • tiffanymak1996
T.T
anonymous
  • anonymous
what happens if we -1/1+0=-1 now what if simply the bottom to -1/1=-1 than what?
tiffanymak1996
  • tiffanymak1996
if it's -1 then the series is convergent, not divergent.
anonymous
  • anonymous
yes..
tiffanymak1996
  • tiffanymak1996
well, the model answer IS divergent.
anonymous
  • anonymous
only if stays at one, -1=-1 if you add one you get 0
tiffanymak1996
  • tiffanymak1996
and that's what frustrates me the most.
tiffanymak1996
  • tiffanymak1996
ok...
anonymous
  • anonymous
so I am taking the answer can't be 0 for the final answer?
tiffanymak1996
  • tiffanymak1996
no, it's not that it can't be 0, just that how do u suddenly +1 after the limit?
anonymous
  • anonymous
Because the +1 falls within the limit.
tiffanymak1996
  • tiffanymak1996
oh, ok, thx, i think i get it now. :)
anonymous
  • anonymous
welcome ...
tiffanymak1996
  • tiffanymak1996
sry for bothering you so long...

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