Question

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Answer 1

So this is a limit problem? Not differentiation.
Ok ok

Answer 2

L'Hopital's rule will involve differentiation :) Yes, good call.

Answer 3

\[\large\rm y=\lim_{x\to\infty}\left(1+\frac1x\right)^x\]If we try to evaluate this using direct substitution, we get the indeterminate form \(\large\rm 1^{\infty}\). So ummm, ok ya your steps look good so far.

Answer 4

Sure, if you'd like :)\[\large\rm \ln y = \lim_{n\to\infty}\frac{\ln\left(1+\frac1n\right) }{\frac1n}\]

Answer 5

\[\large\rm \ln y = \lim_{n\to\infty}\frac{\ln\left(1+t\right) }{t}\]So what happens with our limit?
Where is t going as n grows?

Answer 6

infity over infity, so indeterminate form and use L'Hospital again?

Answer 7

No, I mean the thing written UNDER the limit notation, see it? :)
n is approaching infinity,
what's happening with t,
not with the entire expression.

Answer 8

\[\sqrt{2}\]

Answer 9

Zenmo: I'm curious: why doesn't your math problem appear in your post? And what about the instructions for this problem (find what...?... using log diff'n.)?

Answer 10

My comment (above) pertained to the previous problem that you posted, but my questions are still valid.

Answer 11

\[\sum_{n=1}^{\infty}(1+\frac{ 1 }{ n })^{n^2}\] Original Problem. Yes, give me time to type it all out and you will see my point (of question).

Answer 12

Your "Determine whether the series is absolutely convergent, conditionally convergent, or divergent." does not apply to the expression you typed in:\[\lim_{n \rightarrow \infty}(1+\frac{ 1 }{ n })^n\]

Answer 13

I'd suggest you start over. Make sure that your instructions and your math statement match.

Answer 14

hi mathmale >:)

Answer 15

\[\left| a _{n} \right|=a _{n}=(1+\frac{ 1 }{ n })^{n^2}\]
Using "RATIO TEST,"
\[\lim_{n \rightarrow \infty}\sqrt[n]{(1+\frac{ 1 }{ n })^{n^2}}=\lim_{n \rightarrow \infty}(1+\frac{ 1 }{ n })^{\frac{ n^2 }{ n }}=\lim_{n \rightarrow \infty}(1+\frac{ 1 }{ n })^n\]

Answer 16

Now, I need to find the limit (which I need to apply differentiation).

Answer 17

Did you understand my last comment?
We didn't finish making our substitution,\[\large\rm \ln y = \lim_{\color{red}{n\to\infty}}\frac{\ln\left(1+t\right) }{t}\]

Answer 18

Yes, but mathdale wanted to see my question.

Answer 19

ya np :)

Answer 20

does ln(infinity) approaches infinity ?

Answer 21

Yes, but maybe you're still confused.
We're not allowed to plug anything in for t until you fix this notation under the limit.

Answer 22

Oh the format is wrong.

Answer 23

\[\ln y = \lim_{t \rightarrow \infty}\frac{ \ln(1+t) }{ t }\]

Answer 24

\(\large\rm n\to\infty\) so then if \(\large\rm t=\frac1n\), we have \(\large\rm t\to?\)

Answer 25

No, but good guess! :)

Answer 26

Notice that t is the flip of n, right?

Answer 27

t goes to infinity?

Answer 28

(This is one of the reasons I don't like the substitution, hehe)

Answer 29

No. As n goes to infinity, the reverse of that will be, t goes toward zero.

Answer 30

\[\large\rm n\to\infty:\qquad \frac{1}{n}\to0\]

Answer 31

Yea, with substitution it would be infinity over infinity. Without sub, it will be 0 over 0. Yes?

Answer 32

Yes, good call :D
You're missing some of the important details,
but it will still give us the indeterminate form that we're looking for.\[\large\rm \ln y = \lim_{t\to0}\frac{\ln\left(1+t\right) }{t}\]

Answer 33

It's giving us the indeterminate form that we need in order to be ALLOWED to apply L'Hopital's Rule: 0/0.

Answer 34

ahh ... yes

Answer 35

Yea, forgot that the limit as "some variable" need to --> 0 instead of infinity when using LH.

Answer 36

No fancy quotient rule or anything like that, which is nice.
We differentiate the top and bottom `separately`.

Answer 37

Remember derivative of your log function? :d

Answer 38

\[\ln(1+t) = \frac{ 1 }{ t }\]

Answer 39

Denominator becomes 1, so 1 / t after differentiating?

Answer 40

\[\rm \frac{d}{dt}\ln(1+t)=\frac{1}{1+t}\]

Answer 41

Oh yea, gotcha

Answer 42

\[\large\rm \ln y = \lim_{t\to0}\frac{\ln\left(1+t\right) }{t}\]Hmm ok that cleans things up pretty nicely,\[\large\rm \ln y = \lim_{t\to0}\frac{1}{1+t}\]

Answer 43

So what next?
Maybe direct substitution?

Answer 44

Sorry, post bugged out earlier. I got the problem now. Thanks zepdrix :)