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$$\lim_{n \to \infty} n\bigg(1-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg)$$

If I write expansion of $\bigg(1+\dfrac{1}{n}\bigg)^{n}$ it was equal to expansion of $e$ so $n-n=0$. Is limit is zero ?

Edit: Uploading screen shot of my response. They marked it correct and I don't know if answer key is wrong or not ? Please can anyone say for sure like with $100$ percent surety if answer given is wrong. enter image description here

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    $\begingroup$ Are you sure that you posted a right problem? $\endgroup$ – Michael Rozenberg Feb 10 at 14:12
  • $\begingroup$ I canceled e with $(1+\frac{1}{n})^n$ and editied . $\endgroup$ – Daman deep Feb 10 at 14:14
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    $\begingroup$ Also, delete $n$ in the title $\endgroup$ – Michael Rozenberg Feb 10 at 14:18
  • $\begingroup$ If you add one term to the expansion, you should get $\frac{1}{2}-\frac{11}{24 n}+\frac{7}{16 n^2}+O\left(\frac{1}{n^3}\right)$ which for $n=10$ gives $\frac{2201}{4800}\approx 0.458542$ $\endgroup$ – Claude Leibovici Feb 20 at 10:18
  • $\begingroup$ I don't see your response. $\endgroup$ – zhw. Feb 20 at 16:29
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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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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 illustation, 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 deep Feb 20 at 10:47
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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 at 14:40
  • $\begingroup$ Edited my post please have a look. $\endgroup$ – Daman deep Feb 20 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 deep Feb 20 at 10:47
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Note that $$\lim_{n\to \infty}1-\frac{n}{e}\left(1-\frac{1}{n}\right)^n=-\infty$$

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    $\begingroup$ I think the limit in the body is the correct one. $\endgroup$ – Botond Feb 10 at 14:40

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