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Is the sequence $\left\{\ln\left((1+\frac1n)^n\right)\right\}_{n=1}^{\infty}$ convergent or divergent?.

I tried to solve it by L Hospital's rule and arrived at 0...implying it is convergent..is it? If it is right then is there an alternate method?

user445027
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  • Welcome to MSE! Please add more details to the question, in particular where you encountered this problem and what you have tried so far. Otherwise the community will not be very willing to help you. –  Aug 15 '18 at 09:11
  • I found this problem in Thomas Calculus...I tried l hospital's rule and arrived at 0...which means it is convergent – user445027 Aug 15 '18 at 09:26
  • these details must go into the question, and not in the comments. Moreover, if you applied l’Hospital’s rule to get a precise answer, then what exactly is your question? –  Aug 15 '18 at 09:30
  • Just making if was right, if so, is there any. Alternate method for ir – user445027 Aug 15 '18 at 09:37
  • i see...that is not an issue, all I’m reiterating is that all these details must go into the body of the question itself. Providing your work encourages the community to help you and also makes our work easier. There are also specific tags like [tag:proof-verification] and [tag:alternative-proofs] that you can add to your question if that’s what you’re really asking. Try not to just place a problem statement in the question, that is not considered good practice here. –  Aug 15 '18 at 09:40
  • Is it fine now? – user445027 Aug 15 '18 at 09:43
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    I would be happy to see how you applied L'Hospital's rule as well, just to be sure. After all, solving it by L'Hospital to arrive at $0$ is pretty unclear. What arrives at $0$? But I'll vote to reopen the question now anyway. Hopefully the community should reopen it soon. But I encourage you to provide your work in detail. –  Aug 15 '18 at 09:45
  • @user445027 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Sep 06 '18 at 23:56

1 Answers1

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HINT

Recall that by standard limit

$$\left(1+\frac1n\right)^n\to e$$

As an alternative we have by $x=\frac 1n \to 0$

$$\ln \left(1+x\right)^{\frac1x}=\frac{\ln (1+x)}{x}$$

which is also a standard limit and which, as an alternative, can be solved by l'Hopital rule.

user
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  • How about applying l hospital's rule....then it will come out to be 0 – user445027 Aug 15 '18 at 07:47
  • @user445027 We don't need that, the given limit is a well known limit which should be proved a part. Once we have that by continuity lim_{n\to infty} $$\ln \left(1+\frac1n\right)^n=\ln e =1$$ – user Aug 15 '18 at 07:57
  • @user445027 I've added something to solve it by l'Hopital. Anyway I suggest to learn the standard limits and how to prove them without l'Hopital if you really want to learn to calculate limits effectively. – user Aug 15 '18 at 08:01
  • @ gimusi: correct is $\ln \left(1+\frac1n\right)^n \to\ln e =1$ (not "$=$"): – Fred Aug 15 '18 at 08:02
  • @gimusi how can this be solved without l hospital's rule? – user445027 Aug 15 '18 at 08:03
  • @Fred Yes of course sorry for tha typo. Thanks! – user Aug 15 '18 at 08:03
  • @user445027 The proof is not trivial without l'Hopital but it is really instructive. Notably we need to show that the sequence $\left(1+\frac1n\right)^n$ is increasing and bounded above (<3) then for the monotonic sequence theorem the limit exists. Take a look here https://math.stackexchange.com/questions/1867964/monotonicity-of-the-sequence-a-n-where-a-n-left-1-frac1n-right?noredirect=1&lq=1 and https://math.stackexchange.com/questions/1185660/prove-1-frac1nn-is-bounded-above?noredirect=1&lq=1. – user Aug 15 '18 at 08:09
  • Take also a look here http://www.milefoot.com/math/calculus/limits/LimitDefinitionOfE10.htm – user Aug 15 '18 at 08:14
  • @Fred I just note that I wrote correctly that $\lim_{n\to \infty} a_n=\ln e=1$ but there was a problem in the editing. – user Aug 15 '18 at 09:56