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Compute$$\lim_{n\to\infty}\left(1+\frac{2}{n^2}\right)^n $$

I dont know how to do it without using continuity of exponential function

I mean I cannot do this: $$\lim_{x\to\infty}a_n =a \wedge \lim_{x\to\infty}b_n =b\Rightarrow \lim_{x\to\infty}{a_n}^{b_n} =a^b$$

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  • $\begingroup$ Do you mean $\lim_{n\to\infty}$ rather than $\lim_{x\to\infty}$? $\endgroup$ – Ben Sheller Dec 8 '16 at 20:27
  • $\begingroup$ Yea, sorry my bad $\endgroup$ – UfmdFkiF Dec 8 '16 at 20:30
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    $\begingroup$ $$1\leqslant1+\frac2{n^2}\leqslant e^{2/n^2}\implies1\leqslant\left(1+\frac2{n^2}\right)^n\leqslant e^{2/n}\to1$$ $\endgroup$ – Did Dec 8 '16 at 20:43
  • $\begingroup$ "I dont know how to do it without using (the) continuity of (the) exponential function" What do you call "using (the) continuity of (the) exponential function" here? $\endgroup$ – Did Dec 8 '16 at 20:53
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Use Bernoulli:

$$\left(1+\frac{2}{n^2}\right)^n= \frac{1}{\left(\frac{n^2}{n^2+2}\right)^n}= \frac{1}{\left(1-\frac{2}{n^2+2}\right)^n}$$

And by Bernoulli $$\left(1-\frac{2}{n^2+2}\right)^n \geq 1-\frac{2n}{n^2+2}=\frac{n^2-2n+2}{n^2+2}$$

Therefore $$1 \leq \left(1+\frac{2}{n^2}\right)^n \leq \frac{n^2+2}{n^2-2n+2}$$

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  • $\begingroup$ I'm not sure how well-known "Bernoulli" is in this context. For completeness, then, "Bernoulli" refers to Bernoulli's inequality (en.wikipedia.org/wiki/Bernoulli's_inequality), which is the fact that $(1+x)^r \ge 1+rx$, for positive integer $r$ and real $x$ with $x \ge -1$. $\endgroup$ – Michael Lugo Dec 8 '16 at 21:53
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Hint: the expression equals $[(1+2/n^2)^{n^2}]^{1/n}.$

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  • $\begingroup$ I have no idea how it would help. $\endgroup$ – UfmdFkiF Dec 8 '16 at 20:32
  • $\begingroup$ Except you meant this $[(1+2/n^2)^{n^2/2}]^{2/n}$ but as I said I can't do like that $\endgroup$ – UfmdFkiF Dec 8 '16 at 20:34
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    $\begingroup$ I know that this kind of trick is routinely suggested on the site as if it was elementary and that it is regularly upvoted and chosen as the canonical approach but it always surprises me that users can "swallow" the assertion that $$\lim x_n= x\implies \lim(x_n)^{2/n}=\lim x^{2/n}=1$$ as if it was trivial that $$\lim x_n=x\ \&\lim f_n(x)=\ell\implies\lim f_n(x_n)=\ell$$ And suddenly the implication does not seem so obvious, does it? $\endgroup$ – Did Dec 8 '16 at 20:47
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    $\begingroup$ @TheMeff You can, and even, $(1+1/x)^x\leqslant e$ for every $x>0$. $\endgroup$ – Did Dec 8 '16 at 21:05
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    $\begingroup$ Ouch yourself, there is no misleading approach. Geez, this is how you spend your time today? $\endgroup$ – zhw. Dec 8 '16 at 21:48
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You can consider the limit of the function (as opposed to sequence) $$ f(x)=\left(1+\frac{2}{x^2}\right)^{\!x} $$ or, better yet, the limit of its (natural) logarithm: $$ \lim_{x\to\infty}\log\left(1+\frac{2}{x^2}\right)^{\!x}= \lim_{x\to\infty}x\log\left(1+\frac{2}{x^2}\right)= \lim_{t\to0^+}\frac{\log(1+2t^2)}{t} $$ after the substitution $x=1/t$.

This is the derivative at $0$ of $g(t)=\log(1+2t^2)$ and $$ g'(t)=\frac{4t}{1+2t^2} $$ so $g'(0)=0$. Hence $$ \lim_{x\to\infty}\log f(x)=0 $$ and therefore $$ \lim_{x\to\infty}f(x)=e^0=1 $$ If the limit of the function exists, it is the same as the limit of the sequence.

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  • $\begingroup$ If one looks carefully at your answer, one sees you did not actually prove that $g'(0)=0$, only that $g'(t)\to0$ when $t\to0$ -- which is not the result one needs. $\endgroup$ – Did Dec 8 '16 at 21:31
  • $\begingroup$ @Did Sorry? I'm under the impression that $g$ is a $C^\infty$ function. Don't you trust the formulas for the derivatives of the elementary functions? $\endgroup$ – egreg Dec 8 '16 at 21:33
  • $\begingroup$ What? If you intend to reply to my comment, please read what I actually wrote in said comment. Another hint: rather amazingly, my comment addresses what is written in your question, not the existence or inexistence of other arguments I could provide myself to write down a complete proof. $\endgroup$ – Did Dec 8 '16 at 21:37
  • $\begingroup$ @Did Nowhere I say nor use that $\lim_{t\to0}g'(t)=0$. Do you agree that $g'(0)=\lim\limits_{t\to0}\frac{g(t)-g(0)}{t-0}=\lim\limits_{t\to0}\frac{\log(1+2t^2)}{t}$? Do you agree that for every $t$, $g'(t)=4t/(1+2t^2)$? $\endgroup$ – egreg Dec 8 '16 at 21:56
  • $\begingroup$ Upon rereading, you are right. I somewhat misread what you wrote, and saw problems where there is none. Sorry about the noise (and +1 for your answer). $\endgroup$ – Did Dec 9 '16 at 8:25
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Just another way.

Consider $$A=\left(1+\frac{2}{n^2}\right)^n\implies \log(A)=n\log\left(1+\frac{2}{n^2}\right)$$ Now, since $n$ is large, use Taylor series a round $x=0$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)\implies \log\left(1+\frac{2}{n^2}\right)=\frac{2}{n^2}-\frac{2}{n^4}+O\left(\frac{1}{n^6}\right)$$ $$\log(A)=n\log\left(1+\frac{2}{n^2}\right)=\frac{2}{n}-\frac{2}{n^3}+O\left(\frac{1}{n^5}\right)$$ Now, using $A=e^{\log(A)}$ an Taylor again $$A=1+\frac{2}{n}+\frac{2}{n^2}-\frac{2}{3 n^3}-\frac{10}{3 n^4}+O\left(\frac{1}{n^5}\right)$$ which shows the limit and how it is approached.

All of the above was done for infinitely large values of $n$. However, it gives "good" results far even small values of $n$. For example, using $n=10$, the exact resulut would be $$A=\left(1+\frac{1}{50}\right)^{10}=\frac{119042423827613001}{97656250000000000}\approx 1.21899$$ while the approximation would lead to $\frac{1219}{1000}=1.21900$.

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