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