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Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$

I'm pretty sure it goes to zero since $(n+1)^n > n^n$ but when I input large numbers it goes to $0.36$.

Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$.

So how can I find this limit?

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I have seen $4$ answers in less than $4$ minutes.. but i guess the question is "How"?? He might be expecting "How" for $\lim_{n\rightarrow \infty} (1+\frac{1}{n})^n=e$.... –  Praphulla Koushik Dec 7 '13 at 12:33
    
No I know this identity, I just didn't see the 1/e. –  GinKin Dec 7 '13 at 12:35
    
Then there is no problem i guess... –  Praphulla Koushik Dec 7 '13 at 12:35
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This limit is somewhat similar: math.stackexchange.com/questions/568268/… See also here: math.stackexchange.com/questions/358830/… –  Martin Sleziak Dec 7 '13 at 12:37
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9 Answers 9

up vote 47 down vote accepted

$$\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n = \lim_{n\to \infty}\frac {1}{\left(\frac{n+1}{n}\right)^n}=\lim_{n\to \infty}\frac {1}{\left(\frac nn+\frac{1}{n}\right)^n}=\lim_{n\to \infty}\frac {1}{\left(1+\frac{1}{n}\right)^n}$$ We know that: $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ $$\color{red}{\boxed{\displaystyle\color{black}{\quad\therefore\quad\lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n=\frac 1e\quad}}}$$


And for the intuitive part: $$\begin{array}{|c|c|} \hline \text{Value of }\color{black}{n} & \left(\frac{\color{black}{n}}{\color{black}{n}+1}\right)^{\color{black}{n}} \\ \hline 1 & 1.\overline6 \\ 2 & 0.562 \\ 3 & 0.512 \\ 4 & 0.482 \\ 5 & 0.462 \\ 6 & 0.448 \\ 7 & 0.438 \\ 8 & 0.430 \\ 9 & 0.424 \\ 100 & 0.373 \\ \hline \end{array}$$ And: $$\frac 1e\approx0.3678794...$$ Sure enough, we are approaching $1/e$ as $n$ tends to $\infty$.

Furthermore, you could think of it that way:

As $n$ gets really really large, $1/n$ will be very very small.

As $n$ gets really really large, the coefficients and number of terms of $(a+b)^n$ will also get really large. That's why even if $1/n\approx0$ when $n$ gets big, the expression $\left(1+\frac{1}{n}\right)^n$ will approach a number greater than one, and who has to do with the amount of terms and coefficients present in $(a+b)^n$, and that number is simply $e$. To show more what I mean, look at this example: $$\begin{align}(1+\color{blue}{0.1})^{10}=&1^{10}+\color{green}{10} \cdot1^9 (\color{blue}{0.1})+\color{green}{45}\cdot1^8 (\color{blue}{0.1})^2+\color{green}{120} \cdot1^7 (\color{blue}{0.1})^3+\color{green}{210} \cdot1^6 (\color{blue}{0.1})^4\\&+\color{green}{252}\cdot 1^5 (\color{blue}{0.1})^5+\color{green}{210}\cdot 1^4 (\color{blue}{0.1})^6+\color{green}{120}\cdot 1^3\cdot (\color{blue}{0.1})^7+\color{green}{45}\cdot 1^2 (\color{blue}{0.1})^8\\&+\color{green}{10}\cdot 1 (\color{blue}{0.1})^9+(\color{blue}{0.1})^{10}\\\,\\\approx&2.5937424601\end{align}$$

Note: The digit $2$ present in the mathematical constant $e$ comes from:

  • We have: $\left(1+\frac{1}{n}\right)^n\,\,\text{(1)}$ and $1^n=1\quad\text{where $n\in\Bbb R$}$. That's why there will always be a $1$ left no matter how large $n$ is when we expand the expression $\text{(1)}$.

  • From the binomial theorem, the second coefficient in the expanded form of $(a+b)^n$ is just $n$. And therefore the second term in the expansion of $\left(1+\frac{1}{n}\right)^n$ will be: $n\times1^{n-1}\times\frac1n$ which is equal to $1$.

But in the limit the OP asks about, we are dividing $1$ by $\left(1+\frac{1}{n}\right)^n$, so it is natural that we get at the end $1/e$.

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+0.75 for the awesome answer. +0.25 for the colorful post! –  Ahaan Rungta Dec 7 '13 at 23:01
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$$ \left(\dfrac{n}{n+1}\right)^n = \frac {1}{\left(\dfrac{n+1}{n}\right)^n}=\frac {1}{\left(1+\dfrac{1}{n}\right)^n}$$

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The most elegant! –  Mitsos Dec 7 '13 at 12:34
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Thanks! And also to "Martin Sleziak" for editing. –  randuser Dec 7 '13 at 12:40
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And also to "Adobe" for re-editing! And to myself for re-editing! –  Ahaan Rungta Dec 7 '13 at 23:01
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Yes, of course! –  randuser Dec 8 '13 at 5:57
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Did you know that $\lim_{n\rightarrow \infty} (1+1/n)^n =e$? This you can use to find the limit.

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I came at this problem a little differently than the other answers. We want to find $$L = \lim\limits_{n \to \infty} \left(\frac{n}{n+1}\right)^n.$$ Since the expression inside the limit is always positive, we should be able to take the log of the limit, then solve it (we just have to remember to transform back at the end). $$\begin{align} \log(L) = & \lim\limits_{n \to \infty} n \log \left( \frac{n}{n+1} \right) \\ = & \lim\limits_{n \to \infty} n \log \left( 1 - \frac{1}{n+1} \right). \end{align}$$ Since we're taking the limit of $\log(1 + x)$ as $x \to 0$, what we really care about is the first order behavior of $\log(1 + x)$, so we take the Taylor series: $$\log(1 + x) = x + O(x^2),$$ where $O(x^2)$ indicates higher-order terms (second-order and greater). So $\log(1 + x) \approx x$ for small $x$ ($x \approx 0$). Substituting back into our limit: $$\begin{align} \log(L) = & \lim\limits_{n \to \infty} n \left( -\frac{1}{n+1} \right) \\ = & -1. \end{align}$$ Therefore $L = e^{-1} = \frac{1}{e}$.

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Hint:

$$\left(\frac{n}{n+1}\right)^n=\left(\frac{n+1-1}{n+1}\right)^n=\left(1-\frac{1}{n+1}\right)^n$$

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Thank you, Martin! –  Salech Alhasov Dec 7 '13 at 12:39
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${\lim_{n\to \infty}{(\frac{n}{n+1})}^{n}}={\lim_{n\to \infty}{({1-{\frac{1}{n+1}}})}^n}=e^{\lim_{n\to \infty} \frac{-1}{n+1}*n}=e^{-1}$ where 3$^{rd}$ step is basic process for limit of form $1^{\infty}$ .So limit is $e^{-1}$

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My hint: $$\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^n}\sim \frac{1}{e}$$

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Just write this as $(\frac{n+1}{n})^{-n}$ and then it is a standard limit ie answer will be $1/e$

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \pars{n \over 1 + n}^{n} = \pars{1 + {1 \over n}}^{-n} = \expo{-n\ln\pars{1 + 1/n}}\stackrel{n\ \gg\ 1}{\LARGE\sim} \expo{-n\pars{1/n}} = \expo{-1} $$

$$ \color{#0000ff}{\large\lim_{n \to \infty}\pars{n \over 1 + n}^{n}} = \color{#0000ff}{\large{1 \over e}} $$

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Already posted, in a much more convincing version, by at least one other user. –  Did Dec 8 '13 at 8:02
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protected by Community Dec 7 '13 at 23:25

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