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I tried to solve this limit: $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n$.

My approach was to re-write it as $\lim\limits_{n \to +\infty} \bigg(\dfrac{n}{n+2} + \dfrac{1}{n+2}\bigg)^n$, and since $\dfrac{n}{n+2}$ tends to 1 and $\dfrac{1}{n+2} \sim \dfrac{1}{n}$ as $n \to +\infty$, I figured the solution would be $e$, as $\lim\limits_{n \to +\infty} \bigg(1+\dfrac{1}{n}\bigg)^n = e$.

I suppose I've done something wrong, since by plotting the function I noticed the solution is $\dfrac{1}{e}$.

Where is my error?

mell_o_tron
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    You can't replace one expression by another unless they are equal. Your error is that $n/(n+1)\neq 1$ and yet you replace it by $1$. Same mistake in replacing $1/(n+2)$ with $1/n$. Never fall in this trap even if you get the right answer by fluke. Also I don't understand what is the origin of such heuristics to evaluate limits? Limits are evaluated using theorems meant to evaluate limits. – Paramanand Singh May 11 '21 at 10:54
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    Can you relate the well known limit $(1+(1/n))^n\to e$ with the limit in question?? Do you know that if a sequence tends to a limit $L$ so does every subsequence? – Paramanand Singh May 11 '21 at 11:02
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    Hint: $$ \left( {\frac{{n + 1}}{{n + 2}}} \right)^n = \frac{1}{{\left( {1 + \frac{1}{{n + 1}}} \right)^n }} = \frac{1}{{\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} }}\left( {1 + \frac{1}{{n + 1}}} \right) $$ – Gary May 11 '21 at 11:03

6 Answers6

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The error lies in assuming that, since $\lim_{n\to\infty}\frac n{n+2}=1$ and since $\frac1{n+2}\sim\frac1n$, then $\lim_{n\to\infty}\left(\frac n{n+2}+\frac1{n+2}\right)^n=\lim_{n\to\infty}\left(1+\frac1n\right)^n$. To see why it doesn't work, consider the limit$$\lim_{n\to\infty}n\left(\frac n{n+1}-1\right).$$It is equal to $1$, right?! However, by your argument, since $\lim_{n\to\infty}\frac{n+1}n=1$, it should be equal to$$\lim_{n\to\infty}n(1-1)=0.$$

José Carlos Santos
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$$ \lim_{n\rightarrow \infty}(\frac{n+1}{n+2})^n=\lim_{n\rightarrow \infty}\frac{1}{(\frac{n+2}{n+1})^n}=\lim_{n\rightarrow \infty}\frac{1+\frac{1}{1+n}}{(1+\frac{1}{1+n})^{n+1}}=\frac{1}{e} $$

lemon
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HINTS

  • $\dfrac{n+2}{n+1}=\dfrac{(n+1)+1}{n+1}=1+\dfrac{1}{n+1}$

or

  • $\dfrac{n+1}{n+2}=\dfrac{(n+2)-1}{n+2}=1+\dfrac{1}{-(n+2)}$

And recall that $(1+1/a_n)^{a_n}\to e$ whenever $a_n\to0$.

Tito Eliatron
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Hint:

$\dfrac{(1+1/n)^n}{(1+2/n)^n}.$

Limit of numerator and

denominator $(\not=0)$ exist.

Use: $\lim_{n \rightarrow \infty} (1+x/n)^n=e^x$, $x$ real.

Peter Szilas
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Try replacing the variable $n$ with $x=\frac{1}{n}$: $$\lim_{n \to +\infty} (\frac{n+1}{n+2})^n = \lim_{x \to 0} (\frac{1+x}{1+2x})^\frac{1}{x} = exp(\lim_{x \to 0}\frac{1}{x} (\frac{1+x}{1+2x}-1))=exp(\lim_{x \to 0}\frac{-1}{1+2x})=exp(-1)=\frac{1}{e}$$

Upayan De
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You must know that $\left(1+\frac{1}{n}\right)^n \xrightarrow{n\rightarrow \infty} e$. So, we are going to play wiht it.

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

Now changing $m=n+1$ we have:

$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)$.

Now taking limits when $n$ tends to $+\infty$, as $m=n+1$, is the same as taking limits when $m$ tends to $+\infty$. Then, as $\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty}1$:

$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty} \frac{1}{e}$.

GoRza
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