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I try to find

$$\lim_{n\rightarrow\infty}\bigg(1+\frac{4}{n+8}\bigg)^n$$

So I did a ratio test,

$$\lim_{n\rightarrow\infty}\frac{\bigg(1+\frac{4}{n+8}\bigg)^n}{\bigg(1+\frac{4}{n+9}\bigg)^{n+1}}$$

$$\lim_{n\rightarrow\infty}\frac{\bigg(1+0\bigg)^n}{\bigg(1+0\bigg)^{n+1}}$$

$$\lim_{n\rightarrow\infty}\frac{\bigg(1\bigg)^n}{\bigg(1\bigg)^{n+1}}=1$$

But this is wrong.

What is the right way to solve this?

Thanks

Luthier415Hz
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    Use $\lim(1+\frac{1}{n})^n=e$. Let $n+8=4x$. – Abel Wong Oct 19 '22 at 09:35
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    @AbelWong Rather use $\lim(1+\frac4{n+8})^{n+8}=\lim(1+\frac4m)^m=e^4,$ and $\lim(1+\frac4{n+8})^8=1.$ – Anne Bauval Oct 19 '22 at 09:39
  • Simplify the ratio before taking the limit. – Rócherz Oct 19 '22 at 09:41
  • @Rócherz how? . – Luthier415Hz Oct 19 '22 at 09:43
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    Anyway, your step $\lim_{n\to\infty}\frac{\left(1+\frac4{n+8}\right)^n}{\left(1+\frac{4}{n+9}\right)^{n+1}}=\lim_{n\to\infty}\frac{\left(1+0\right)^n}{\left(1+0\right)^{n+1}}$ was not correct. With such an erroneous "method", you could derive that $\lim_{n\to\infty}\left(1+\frac4{n+8}\right)^n=1,$ whereas in reality, this limit is $e^4.$ – Anne Bauval Oct 19 '22 at 09:44

2 Answers2

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$\lim_{n\to∞}( 1 + \frac{4}{n+8} )^n$

$=\lim_{n\to∞}( 1 + \frac{4}{n+8} )^{(n+8) -8}$

$=\lim_{n\to∞}\frac{( 1 + \frac{4}{n+8} )^{(n+8)}}{(1 + \frac{4}{n+8})^8}$

$= e^4$

Simon Leung
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$$a_n=\left(1+\frac{4}{n+8}\right)^n \quad \implies \quad \log(a_n)=n \log\left(1+\frac{4}{n+8}\right) $$ Let $$\frac{4}{n+8}=\epsilon\implies n=4\frac{1-2 \epsilon }{\epsilon }\implies \log(a_n)=4\frac{1-2 \epsilon }{\epsilon }\log(1+\epsilon)$$ Use the series expansion of $\log(1+\epsilon)$

$$\log(a_n)=4\frac{1-2 \epsilon }{\epsilon }\Big[\epsilon -\frac{\epsilon ^2}{2}+\frac{\epsilon ^3}{3}+O\left(\epsilon ^4\right) \Big]$$ that is to say

$$\log(a_n)=4-10 \epsilon +\frac{16 }{3}\epsilon ^2+O\left(\epsilon ^3\right)$$

Continue with Taylor series $$a_n=e^{\log(a_n)}=e^4\Big[1-10 \epsilon +\frac{166 }{3}\epsilon ^2+O\left(\epsilon ^3\right)\Big]$$

Claude Leibovici
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