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How do you calculate : $$\lim_{n\to \infty}\frac{n^{2}}{\left(2+\frac{1}{n}\right)^{n}}$$

Thanks!

Daniel Fischer
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Shahar Shalev
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  • show that $\lim_{n \to \infty}\left(2+\frac{1}{n}\right)^n=\infty$ – Dr. Sonnhard Graubner Dec 14 '15 at 18:07
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    It's very obviously zero, exponential functions with base greater than 1 rise faster than any power. – orion Dec 14 '15 at 18:08
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    In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – gebruiker Dec 14 '15 at 18:22
  • We can bound the terms by $n^2/2^n $, which is easier to work with. – bilaterus Dec 14 '15 at 19:03

5 Answers5

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Let $a_n = \dfrac{n^{2}}{\left(2+\frac{1}{n}\right)^{n}}$.

Then $\sqrt[n]{a_n} = \dfrac{(\sqrt[n]{n})^{2}}{2+\frac{1}{n}} \to \dfrac12 < 1$.

By the root test, $\sum a_n$ converges and so $a_n \to 0$.

lhf
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  • See http://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 for a proof that $\sqrt[n]{n}\to1$. – lhf Dec 14 '15 at 18:27
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HINT

What happens if you replace $(2+1/n)^n$ by $2^n$?

gt6989b
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  • formally, you cannot just replace it – Kamil Jarosz Dec 14 '15 at 18:12
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    @kamil09875 this is a hint, not an answer -- the OP is invited to think about his problem himself. But if you ensist, one transforms $$ \frac{n^2}{(2+1/n)^n} = \frac{n^2}{2^n} \left(\frac{2}{2+1/n}\right)^n $$ and bounds the right term by $1$ – gt6989b Dec 14 '15 at 18:17
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$n=\cfrac{x}{2}$ $$\underset{n\rightarrow\infty}{\lim}\cfrac{n^{2}}{\left(2+\dfrac{1}{n}\right)^{n}}=\underset{x\rightarrow\infty}{\lim}\cfrac{\dfrac{x^{2}}{4}}{\left(2+\dfrac{2}{x}\right)^{\frac{x}{2}}}=\underset{x\rightarrow\infty}{\lim}\cfrac{\dfrac{x^{2}}{4}}{2^\frac{x}{2}\sqrt{\left(1+\dfrac{1}{x}\right)^{x}}}=\underset{x\rightarrow\infty}{\lim}\cfrac{\dfrac{x^{2}}{4}}{2^\frac{x}{2}\sqrt{e}}=0$$

Isura Manchanayake
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$$\lim_{n\to\infty}\frac{n^2}{(2+\tfrac1n)^n}=\lim_{n\to\infty}\left(\frac{(2+\tfrac1n)^n}{n^2}\right)^{-1}$$

Write $n^2$ as $(\sqrt[n]{n^2})^n$:

$$\lim_{n\to\infty}\left(\frac{(2+\frac1n)^n}{(\sqrt[n]{n^2})^n}\right)^{-1}=\lim_{n\to\infty}\left(\frac{2}{(\sqrt[n]{n})^2}+\frac{1}{n(\sqrt[n]{n})^2}\right)^{-n}=\left[\left(\frac{2}{1}+\frac{1}{1\cdot\infty}\right)^{-\infty}\right]=0$$

Because $\sqrt[n]{n}\to 1$

Kamil Jarosz
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Write the denominator as $$\left(2+\frac1n\right)^n=2^n\left(1+\frac{1/2}{n}\right)^n \to +\infty\cdot e^{1/2}=+\infty$$ as $n\to +\infty$. So $$\lim_{n\to +\infty}\frac{n^2}{\left(2+\frac1n\right)^n}=e^{-1/2}\lim_{n\to+\infty}\frac{n^2}{2^n}=0$$ where the last equality is due to L'Hopital's rule, since $(2^n)''=2^n(\ln2)^2$ while $(n^2)''=2$ independent of $n$.

Jimmy R.
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