tiffanymak1996
is this absolutely convergent or conditionally or divergent?
Delete
Share
This Question is Closed
tiffanymak1996
Best Response
You've already chosen the best response.
0
\[\sum_{n=1}^{\infty} (1)^{n}\times e ^{\frac{ 1 }{ n }}\]
Herp_Derp
Best Response
You've already chosen the best response.
0
Divergent...
tiffanymak1996
Best Response
You'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?
Herp_Derp
Best Response
You'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?
tiffanymak1996
Best Response
You've already chosen the best response.
0
but, i thought n=1 under the sigma sign...
godorovg
Best Response
You've already chosen the best response.
1
No, tiffany they are telling right here, it is diverges
jeanlouie
Best Response
You'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
tiffanymak1996
Best Response
You'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
tiffanymak1996
Best Response
You'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
jeanlouie
Best Response
You'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
godorovg
Best Response
You'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?
tiffanymak1996
Best Response
You'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
tiffanymak1996
Best Response
You'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.
tiffanymak1996
Best Response
You'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.
tiffanymak1996
Best Response
You've already chosen the best response.
0
although i know i'm wrong, cause the answer is divergent:(
godorovg
Best Response
You've already chosen the best response.
1
what is the value {1} of this when testing ratio =
tiffanymak1996
Best Response
You've already chosen the best response.
0
when testing ratio without the absolute value i got 1
godorovg
Best Response
You've already chosen the best response.
1
How does the value change to 1 I am lost there?
jeanlouie
Best Response
You'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
jeanlouie
Best Response
You've already chosen the best response.
0
ϵ< 2, sorry
tiffanymak1996
Best Response
You'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
Best Response
You've already chosen the best response.
0
ln both top and bottom
jeanlouie
Best Response
You've already chosen the best response.
0
yeah, ρ = 1, you got it
tiffanymak1996
Best Response
You've already chosen the best response.
0
becomes:\[\lim_{n \rightarrow +\infty} \frac{ \frac{ 1 }{ n+1} }{ \frac{ 1 }{ n } } \]
jeanlouie
Best Response
You've already chosen the best response.
0
no, top or bottom may be negative, we can't use ln here.
tiffanymak1996
Best Response
You'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
Best Response
You've already chosen the best response.
1
Tiffany I think you have the right answer.
tiffanymak1996
Best Response
You've already chosen the best response.
0
i do? but this is like conditionally ; yet the modal answer is divergent...
godorovg
Best Response
You've already chosen the best response.
1
Tiffany 1+1n is not 1 it is 2n
tiffanymak1996
Best Response
You've already chosen the best response.
0
model, sry
godorovg
Best Response
You've already chosen the best response.
1
2/n not 1
godorovg
Best Response
You've already chosen the best response.
1
you see where this is?
tiffanymak1996
Best Response
You'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
tiffanymak1996
Best Response
You've already chosen the best response.
0
is this correct?
godorovg
Best Response
You've already chosen the best response.
1
when sub in n, you still have 1+1 which is 2.
tiffanymak1996
Best Response
You've already chosen the best response.
0
where does the 1+1 com from?
godorovg
Best Response
You'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?
tiffanymak1996
Best Response
You've already chosen the best response.
0
so sorry, it should be 1/(1+(1/n)) not 1/(1+1/n)
godorovg
Best Response
You've already chosen the best response.
1
1/1=1times +1 =1/n
tiffanymak1996
Best Response
You've already chosen the best response.
0
\[\lim_{n \rightarrow +\infty}\frac{ 1 }{ 1+\frac{ 1 }{ n } }\]
godorovg
Best Response
You've already chosen the best response.
1
yes..
tiffanymak1996
Best Response
You've already chosen the best response.
0
so...
\[\lim_{n \rightarrow +\infty} \frac{ 1 }{ 1+\frac{ 1 }{ n } }\]
godorovg
Best Response
You've already chosen the best response.
1
Tiffany can the bottom be 0? or am I missing something here?
tiffanymak1996
Best Response
You've already chosen the best response.
0
the bottom can't be zero, what i did is just divide top and bottom with n
tiffanymak1996
Best Response
You've already chosen the best response.
0
and then:
\[\frac{ 1 }{ 1+\frac{ 1 }{ \infty } }\]
tiffanymak1996
Best Response
You've already chosen the best response.
0
and since \[\frac{ 1 }{ \infty }=0\],
tiffanymak1996
Best Response
You've already chosen the best response.
0
\[\frac{ 1 }{ 1+0 }=1\]
tiffanymak1996
Best Response
You've already chosen the best response.
0
and i did this like 5 times and still don't get why it's divergent...
tiffanymak1996
Best Response
You've already chosen the best response.
0
T.T
godorovg
Best Response
You'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?
tiffanymak1996
Best Response
You've already chosen the best response.
0
if it's 1 then the series is convergent, not divergent.
godorovg
Best Response
You've already chosen the best response.
1
yes..
tiffanymak1996
Best Response
You've already chosen the best response.
0
well, the model answer IS divergent.
godorovg
Best Response
You've already chosen the best response.
1
only if stays at one, 1=1 if you add one you get 0
tiffanymak1996
Best Response
You've already chosen the best response.
0
and that's what frustrates me the most.
tiffanymak1996
Best Response
You've already chosen the best response.
0
ok...
godorovg
Best Response
You've already chosen the best response.
1
so I am taking the answer can't be 0 for the final answer?
tiffanymak1996
Best Response
You'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?
godorovg
Best Response
You've already chosen the best response.
1
Because the +1 falls within the limit.
tiffanymak1996
Best Response
You've already chosen the best response.
0
oh, ok, thx, i think i get it now. :)
godorovg
Best Response
You've already chosen the best response.
1
welcome ...
tiffanymak1996
Best Response
You've already chosen the best response.
0
sry for bothering you so long...