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I'm working on a practice exam for my masters quals and I am having difficulties with the following limit. According to wolfram alpha, it's value is 1/2. Does anyone know how to find this limit?

$$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$

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  • $\begingroup$ Can you use L'Hospital's Rule? $\endgroup$
    – graydad
    Jan 6, 2015 at 21:38
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    $\begingroup$ This table may help you study for your exam. It tells you how to transform the indeterminates to easier forms. $\endgroup$ Jan 6, 2015 at 21:39
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    $\begingroup$ Are you sure that you posted a right problem? $\endgroup$ Feb 10, 2019 at 14:12
  • $\begingroup$ I canceled e with $(1+\frac{1}{n})^n$ and editied . $\endgroup$
    – Daman
    Feb 10, 2019 at 14:14
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    $\begingroup$ Also, delete $n$ in the title $\endgroup$ Feb 10, 2019 at 14:18

8 Answers 8

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Just use the fact that $$\left(1+\frac1n\right)^n=e^{n\ln(1+\frac1n)}=e^{n\left(\frac1n-\frac{1}{2n^2}+o(\frac{1}{n^2})\right)}=e\cdot e^{-\frac{1}{2n}+o(\frac1n)}=e\left(1-\frac{1}{2n}+o\Bigl(\frac{1}{n^2}\Bigr)\right),$$and the result follows.

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Hint:

$$\lim \limits_{n\rightarrow \infty} n-\frac{n}{e}\left(1 +\frac{1}{n}\right)^n = \lim \limits_{n\rightarrow \infty} \frac{1-\frac{1}{e}\left(1 +\frac{1}{n}\right)^n}{\frac{1}{n}},$$ and use l'Hopital rule.

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  • $\begingroup$ this answer is better and alot more simple. Good job. $\endgroup$ Jan 6, 2015 at 21:54
  • $\begingroup$ @Bot, thanks for the comment :) $\endgroup$
    – Alex Silva
    Jan 6, 2015 at 21:56
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    $\begingroup$ I don't know if this is better, since "use l'Hopital rule" is where all the work is. $\endgroup$ Jan 6, 2015 at 23:12
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By the L'Hôpital's rule we obtain: $$\lim_{n\rightarrow+\infty}n\left(1-\frac{1}{e}\left(1+\frac{1}{n}\right)^{n}\right)=\lim_{x\rightarrow0^+}\frac{e-(1+x)^{\frac{1}{x}}}{ex}=-\frac{1}{e}\lim_{x\rightarrow0^+}\left(e^{\frac{\ln(1+x)}{x}}\right)'=$$ $$=-\frac{1}{e}\lim_{x\rightarrow0^+}(1+x)^{\frac{1}{x}}\left(\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}\right)=$$ $$=-\frac{1}{e}\lim_{x\rightarrow0^+}(1+x)^{\frac{1}{x}}\lim_{x\rightarrow0^+}\frac{x-(1+x)\ln(1+x)}{x^2+x^3}=$$ $$=-\lim_{x\rightarrow0^+}\frac{1-\ln(1+x)-1}{2x+3x^2}=\lim_{x\rightarrow0^+}\frac{\ln(1+x)}{x}\lim_{x\rightarrow0^+}\frac{1}{2+3x}=\frac{1}{2}.$$

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We have \begin{align} \lim_{n\to\infty}n\bigg[1-\dfrac{n}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg] =& \lim_{n\to\infty}n^2\bigg[\frac{1}{n}-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg] \\ =& \lim_{n\to\infty}n^2\cdot \lim_{n\to\infty}\bigg[\frac{1}{n}-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg] \\ =& \lim_{n\to\infty}n^2\cdot \bigg[\lim_{n\to\infty}\frac{1}{n}-\dfrac{1}{e}\lim_{n\to\infty}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg] \\ =& \lim_{n\to\infty}n^2\cdot \bigg[0-\dfrac{1}{e}e \bigg] \\ =& \lim_{n\to\infty}-n^2 =-\infty \end{align}

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    $\begingroup$ I think the limit in the body is the correct one. $\endgroup$
    – Botond
    Feb 10, 2019 at 14:40
  • $\begingroup$ Edited my post please have a look. $\endgroup$
    – Daman
    Feb 20, 2019 at 10:47
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Let us consider $$\lim_{n \to \infty} n\bigg(1-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg)$$ First, let $$a_n=\bigg(1+\dfrac{1}{n}\bigg)^{n}\implies \log(a_n)=n \log\bigg(1+\dfrac{1}{n}\bigg)$$ that is to say $$\log(a_n)=n\bigg(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right) \bigg)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor $$a_n=e^{\log(a_n)}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$1-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n}=\frac{1}{2 n}-\frac{11}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$n\bigg(1-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg)=\frac{1}{2}-\frac{11}{24 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

For illustration, use your pocket calculator for $n=10$; the exact value is $10-\frac{25937424601}{1000000000 e}\approx 0.458155$ while the above expansion gives $\frac{109}{240}\approx 0.454167$.

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  • $\begingroup$ Edited my post please have a look. $\endgroup$
    – Daman
    Feb 20, 2019 at 10:47
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I'll present you the way I always approach these kinds of problems. You can or should be able to transform this into a rigorous argument. However, if you have any questions, feel free to ask them!

First, we try to find how quick the inside converges to $0$, i.e., how quick does this standard limit converge to $e$. Thereto, we take the logarithm of the stuff inside $$\log\left(\left(1+\frac1n\right)^n\right) = n\log\left(1+\frac1n\right) = n\cdot\left(\frac1n - \frac{1}{2n^2} +O(n^{-3})\right) = 1 -\frac{1}{2n} +O(n^{-2})$$ Hence, $$1-\frac1e\left(1+\frac1n\right)^n = 1-\frac1e\cdot \exp\left(1-\frac1{2n}+O(n^{-2})\right) = 1 - \exp\left(-\frac1{2n}+O(n^{-2})\right)$$ Finally, we need to multiply this with $n$. Can you finish this?

Since you added a bounty. Let me finish this for you. So in the end it boils down by finding the limit $$\lim_{n\to\infty} n\left(1 - \exp\left(-\frac1{2n}+O(n^{-2})\right)\right)$$ Taylor states that $$\exp\left(-\frac1{2n}+O(n^{-2})\right) = 1 - \frac1{2n} + O(n^{-2})$$ And thus, $$\lim_{n\to\infty} n\left(1 - \exp\left(-\frac1{2n}+O(n^{-2})\right)\right) =\lim_{n\to\infty} n\left(\frac1{2n} + O(n^{-2})\right)=\lim_{n\to\infty}\frac12 +O(n^{-1}) =\frac12$$

Does this make sense?

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  • $\begingroup$ Edited my post please have a look. $\endgroup$
    – Daman
    Feb 20, 2019 at 10:47
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In this answer, it is shown, using an elementary extension of Bernoulli's Inequality, that $$ \lim_{n\to\infty}\left(\frac1e\left(1+\frac1n\right)^n\right)^n=e^{-1/2}\tag1 $$ Since $$ \lim_{n\to\infty}\left(1-\frac1{2n}\right)^n=e^{-1/2}\tag2 $$ we have $$ \lim_{n\to\infty}\left[\left(1-\frac1{2n}\right)^n-\left(\frac1e\left(1+\frac1n\right)^n\right)^n\right]=0\tag3 $$ $(1)$, $(2)$, and $\frac{x^n-y^n}{n(x-y)}=\frac1n\sum\limits_{k=1}^nx^{n-k}y^{k-1}$ show that $$ \lim_{n\to\infty}\frac{\left(1-\frac1{2n}\right)^n-\left(\frac1e\left(1+\frac1n\right)^n\right)^n}{n\left[\left(1-\frac1{2n}\right)-\frac1e\left(1+\frac1n\right)^n\right]}=e^{-1/2}\tag4 $$ $(3)$ and $(4)$ imply that $$ \lim_{n\to\infty}n\left[\left(1-\frac1{2n}\right)-\frac1e\left(1+\frac1n\right)^n\right]=0\tag5 $$ Adding $\frac12$ to both sides gives $$ \lim_{n\to\infty}n\left[1-\frac1e\left(1+\frac1n\right)^n\right]=\frac12\tag6 $$

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Hint: try logarithms. $$f(n) - g(n) = \ln\frac{f(n)}{g(n)}$$

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