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This is the second part of a problem that was on my test.

The first part was:

  1. For all $ x > -1$, prove $\log(1+x) - x = \int_{0}^x \frac{(t-x)}{(1+t)^2}dt$

which I did with integration by parts.

The second part was:

  1. Assuming part 1, prove part 2 which is:

for all x in R, $\lim_{n\to\infty} n(\log(1+x/n)-x/n) = 0 $

I can see how this works but, I can't seem to prove it. I tried L'hopital's rule which was becoming too messy, then tried working with the formal definition of a sequence but again I was getting stuck.

Please help.

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Assuming $x$ is positive, the negative case is similar, $$\int_0^{\frac{x}{n}}\frac{\frac{x}{n}-t}{(1+t)^2}dt\leq \int_0^{\frac{x}{n}}\frac{\frac{x}{n}}{(1+t)^2}dt=\frac{x^2}{n^2}\frac{1}{1+\frac{x}{n}}\to 0$$

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Note that

$$n\log(1+x/n)=\log((1+x/n)^n)$$

and we know that $\lim_{n\to\infty}(1+x/n)^n=e^x$

Can you finish it from here?

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  • $\begingroup$ Can the downvoter comment on why you find this answer not useful? $\endgroup$ – vrugtehagel Apr 3 '18 at 19:20
  • $\begingroup$ Its not clear how you expect the question to be completed. Plus I dont like the :"can you finish" comment. $\endgroup$ – Rene Schipperus Apr 3 '18 at 19:22
  • $\begingroup$ The "can you finish it from here" is to make sure I don't just spoil the answer. Clearly, by the way the question's being phrased, the OP is familiar with these concepts and is motivated to solve it, so I avoid just throwing the entire proof in his lap. I think this is the key the OP is missing, hence only giving it this. So "it is not clear" - this is on purpose, as a completely clear answer would not encourage the OP to think for himself. "I don't like ..." is never a reason to downvote. $\endgroup$ – vrugtehagel Apr 3 '18 at 19:25
  • $\begingroup$ Maybe you should take a xanax. $\endgroup$ – Rene Schipperus Apr 3 '18 at 19:27
  • $\begingroup$ proving limn→∞(1+x/n)n=ex using part 2, was actually part 3 of the question. I'll try going from this to part 2, but is there another way to do it? Using maybe the definition of a sequence or something? $\endgroup$ – gunnersFc Apr 3 '18 at 19:33
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Note that by Taylor’s expansion

  • $\log(1+x/n)=\frac x n-\frac12 \frac {x^2}{n^2}+o(n^{-2})$

then

$$n(\log(1+x/n)-x/n)=-\frac12 \frac {x^2}{n}+o(n^{-1})\to 0$$

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