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Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.

I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$

I am trying to go ahead of this step.

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13 Answers 13

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We use the inequality between the geometric mean and the arithmetic mean for the following positive numbers $$ x_{1}=1,~x_{2}=x_{3}=\ldots=x_{n+1}=1+\frac{1}{n}\text{.}% $$ Then $$ \sqrt[n+1]{x_{1}x_{2}\cdots x_{n+1}}<\frac{x_{1}+x_{2}+\ldots+x_{n+1}}{n+1}% $$ (the inequality is strict, since the numbers can't be all equal) translates to $$ \left( 1+\frac{1}{n}\right) ^{\frac{n}{n+1}}<\frac{1+n\left( 1+\frac{1}{n}\right) }{n+1}=1+\frac{1}{n+1}% $$ hence $a_{n}<a_{n+1}$.

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  • $\begingroup$ I like this proof since it uses minimal advanced knowledge :) $\endgroup$ May 31 '21 at 17:42
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$$x_n=\bigg(1+\frac{1}{n}\bigg)^n\longrightarrow x_{n+1}=\bigg(1+\frac{1}{n+1}\bigg)^{n+1}$$ $$\frac{x_{n+1}}{x_{n}}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}}=\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)=\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)$$ $$=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)≥\bigg(1-\frac{n}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg)$$ $$≥^*\frac{1}{1+\frac{1}{n+1}}\bigg(1+\frac{1}{n+1}\bigg)≥1$$ It means that your sequence is increasing.

≥*: $$(n+2)(n^2+n+1)=(n+2)\bigg((n+1)^2-n\bigg)≥(n+1)^3$$

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    $\begingroup$ no problem, to make larger paranthesis, use the command \big(,\big) or \bigg(,\bigg) for extra large ones. You can also use the command \left(,\right) and LateX will automatically scale the size so that it looks nice. $\endgroup$ Jul 7 '12 at 17:56
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    $\begingroup$ I think the inequality $$\left(1-\frac{1}{(n+1)^2}\right)^n\geq 1-\frac{n}{(n+1)^2}$$should be justified by at least mentioning Bernoulli's inequality: $\,\forall -1<x\in \Bbb R\,\,,\,(1+x)^n\geq 1+nx\,$. Anyway, this proof is more elementary. +1 $\endgroup$
    – DonAntonio
    Jul 8 '12 at 2:26
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    $\begingroup$ @DonAntonio: Yes. This proof is more elementary as you noted. Honestly, I knew it from the book Modern Calculus and Analytic Geometry by R. A. Silverman. Thanks $\endgroup$
    – Mikasa
    Jul 8 '12 at 5:10
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    $\begingroup$ @Babak Sorough Thanks. I have chosen this answer. $$Will you please explain$$ $$How\bigg(1−\frac{n}{(n+1)^2}\bigg)\geq\bigg(\frac{1}{1+\frac{1}{n+1}}\bigg)?$$ $\endgroup$ Jul 11 '12 at 8:41
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    $\begingroup$ The analysis flows more easily if we write $$\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}=\left(1+\frac1n\right)\,\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\ge \left(1+\frac1n\right)\,\left(1-\frac{1}{n+1}\right)=1$$Doesn't that seem easier? ;-)) -Mark $\endgroup$
    – Mark Viola
    Aug 20 '16 at 22:17
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The (well known) elementary proof that this sequence is increasing relies on the Bernoulli inequality, which states that, for real $x\ge -1$ and $n\in \mathbb{N}$, $$(1+x)^n \ge 1+nx$$ which can be easily shown by induction. This looks quite inefficient but should not be underestimated. If you know this, then observe that $$\left(1+\frac{1}{n}\right)^{n} > \left(1+\frac{1}{n-1}\right)^{n-1} $$ is equivalent to $$\left( \frac{1+\frac{1}{n}}{1+\frac{1}{n-1}}\right)^{n} > \left(1+\frac{1}{n-1}\right)^{-1} = 1-\frac{1}{n}$$ The lhs is equal to $$ \left(\frac{n^2-1}{n^2}\right)^n = \left(1-\frac{1}{n^2}\right)^n $$ which, according to Bernoulli is $$> 1-\frac{n}{n^2} = 1-\frac{1}{n}$$ which is what was to be shown.

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Take logarithms. You need to compare $n\ln(1+\frac{1}{n})$ to $(n+1)\ln(1+\frac{1}{n+1})$. Because the logarithm is strictly concave, the function (defined for positive $x$) $$\frac{\ln(1+x)}{x}=\frac{\ln(1+x)-\ln(1)}{(1+x)-1}$$ is strictly decreasing (and tends to $1=\ln'(1)$ as $x$ tends to $0$.) Apply this to the striclty decreasing sequence $1/n$ and you get that the sequence $$\frac{\ln(1+1/n)}{1/n}\mathrm{~is~strictly~increasing.}$$ Of course $\frac{\ln(1+1/n)}{1/n}=n\ln(1+\frac{1}{n})$, so, upon exponentiating, $U_n$ is strictly increasing (and tends to $e$.)

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  • $\begingroup$ what do you mean by $ln'(1)$ what is the '? $\endgroup$
    – asdf
    Jul 8 '21 at 11:27
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    $\begingroup$ @asdf it's the derivative of $\ln$. $\endgroup$ Jul 8 '21 at 12:07
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If you expand $(1+\frac1n)^n$ by the binomial theorem, the term involving $1^{n-k}(\frac1n)^k$ is $\binom{n}{k}/n^k$ (I take such a term to exist, and be $0$, in case $k>n$). If one can show that each such term is a monotonically increasing expresion in $n$, then certainly the sum of all terms will be a monotonically increasing expression in $n$ (this involves formally adding up infinitely many expressions, but in comparing $U_n$ and $U_{n+1}$ only finitely many terms are involved, so there is no need to take limits). Now we can write $$ \frac{\binom{n}{k}}{n^k}=\frac1{k!}\cdot\frac{n}{n}\frac{(n-1)}n\cdots\frac{(n-k+1)}n =\frac1{k!}(1-\frac1n)(1-\frac2n)\ldots(1-\frac{(k-1)}n) $$ This expression is zero as long as $n<k$, and beyond that point all factors are positive and either independent of $n$ or increasing expressions in $n$. We may conclude that term $k$ is constant for $k\leq1$, and a weakly increasing function of $n$, strictly increasing as soon as it is nonzero, for $k\geq2$. This proves the result.

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HINT: Differentiate with respect to $n$. Prove this is always increasing.

Side note: Do you know what $a_n$ is?


So we have $f(x) = \left(1+\frac1{n}\right)^n$

$\log(f(x)) = n\log\left(1+\frac1{n}\right)$

$\log(f(x)) = n\log\left(n+1\right) - n\log(n)$

Try differentiating now.

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  • $\begingroup$ i tried its too complicated to conclusively tell it is always positive $\endgroup$
    – raj
    Mar 8 '14 at 19:07
  • $\begingroup$ @raj okay I am adding the differential in. $\endgroup$
    – Guy
    Mar 8 '14 at 19:09
  • $\begingroup$ Actually, it is not quite right to use differentiation. The fact that this sequence is increasing is used for defining Euler's constant, so assuming you don't know that, how can you differentiate? $\endgroup$ Mar 8 '14 at 19:11
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    $\begingroup$ @digital-Ink that depends on your definition of Euler's constant $\endgroup$ Mar 8 '14 at 19:12
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    $\begingroup$ Replace $n$ by $x$ eight times. $\endgroup$
    – Did
    Apr 10 '14 at 21:34
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Let $u_n = (1+1/n)^n$ and $v_n = (1+1/n)^{n+1}$. Then for $n>1$

\begin{align}\frac{u_n}{v_{n-1}}=\bigg(\frac{n+1}{n}\bigg)^n\bigg(\frac{n-1}{n}\bigg)^n =\bigg(1-\frac{1}{n^2}\bigg)^n\\ > 1-\frac{1}{n}=\frac{n-1}{n} \end{align}

So \begin{align}u_n> v_{n-1}\bigg(\frac{n-1}{n}\bigg)=\bigg(1+\frac{1}{n-1}\bigg)^{n-1}=u_{n-1}\end{align}

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Note that the function $f(x) = (1+\frac 1x)^x$ is differentiable. We may find its derivative as follows: $$ \ln(f(x)) = \ln\left((1+\frac 1x)^x\right) = x\ln(1 + \frac 1x)\\ f'(x)/f(x) = \ln\left(1 + \frac 1x\right) + \frac{x}{1+\frac 1x}\cdot \frac{-1}{x^2}\\ f'(x)/f(x) = \ln\left(1 + \frac 1x\right) - \frac{1}{x+1}\\ f'(x) = \left[\ln\left(1 + \frac 1x\right) - \frac{1}{x+1}\right]f(x)\\ f'(x) = \left[\ln\left(\frac {x+1}x\right) - \frac{1}{x+1}\right]f(x)\\ f'(x) = \left[\ln(x+1) - (\ln(x) + \frac{1}{x+1})\right]f(x) $$ Show that the derivative is positive from $x = 2013$ to $x = 2014$.

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    $\begingroup$ What is $\ln$? It is the logarithm with base $\mathrm{e}$. But who is $\mathrm{e}$? Well, it is the limit of $a_n$. But why is $a_n$ convergent? Because it is increasing and bounded. Do you see the chicken and the egg in your approach? $\endgroup$ Mar 8 '14 at 19:14
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    $\begingroup$ @digital-Ink I define $\ln x$ to be the function whose derivative is $1/x$ and whose value at $x = 1$ is $0$. $\endgroup$ Mar 8 '14 at 19:15
  • $\begingroup$ @digital-Ink $e$ has multiple definitions, and though historically your argument would be correct, in this context we can use it. Although you will most definitely get my upvote if you do something without it. $\endgroup$
    – Guy
    Mar 8 '14 at 19:15
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    $\begingroup$ @digital-Ink I define $e$ to be the unique positive number $a$ such that the area under the curve $a^x$ from $a$ to $1$ is $(a-1)$. Oh my. It comes out be equal to this series too? What a coincidence. $\endgroup$
    – Guy
    Mar 8 '14 at 19:17
  • $\begingroup$ @omnomnomnom how can you say $\ln\left(1 + \frac 1x\right) \geq \frac{1}{x+1}$ $\endgroup$
    – raj
    Mar 8 '14 at 19:19
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Wanted to add yet another method to the "catalogue"; I learned this by solving a problem from the book Numbers and Functions: Steps into Analysis by R P. Burn.

First, we have the following lemma:

Let $0 < a < b$. Then $$\frac{b^{n+1}-a^{n+1}}{b-a} < (n+1)b^n$$

This can be proved by noting that $$b^{n+1}-a^{n+1} = (b-a)(b^n + b^{n-1} a + b^{n-2} a^2 + \dots + a^n)$$ and since $0 < a < b \implies 0 < a^n < b^n$, we get $$b^n + b^{n-1} a + b^{n-2} a^2 + \dots + a^n < b^n + b^{n-1} b + b^{n-2} b^2 + \dots + b^n = (n+1)b^n$$.

Now plug in $$a = 1 + \frac{1}{n+1}, \hspace{10pt} b = 1 + \frac{1}{n}$$ into the lemma, and let $$t_n = \left(1 + \frac{1}{n}\right)^n$$ This gives

$$\frac{\overbrace{\left(1+\frac{1}{n}\right)^{n+1}}^{(1+\frac{1}{n})t_n} - \overbrace{\left(1 + \frac{1}{n+1}\right)^{n+1}}^{t_{n+1}}}{\frac{1}{n}-\frac{1}{n+1}} < (n+1)\overbrace{\left(1+\frac{1}{n}\right)^n}^{t_n}$$ which after some straightforward algebraic simplification yields $t_n < t_{n+1}$.

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  • $\begingroup$ I saw a similar proof in a book I was reading. Can you give me an idea on how can one come up with something so elaborate? $\endgroup$
    – eem
    Jul 26 '19 at 17:26
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My two cents (as later similar question closed):

$$\frac{a_{n+1}}{a_n}=\frac{\left( 1+ \frac{1}{n+1} \right)^{n+1}}{\left( 1+ \frac{1}{n} \right)^{n}}=\frac{n+1}{n}\left( 1- \frac{1}{(n+1)^2} \right)^{n+1}>\\ >\frac{n+1}{n}\left( 1- \frac{1}{(n+1)} \right)=1$$

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  • $\begingroup$ I like this answer best. It's the simplest. On the other hand, from the accepted answer I learned that $(n+2)(n^2+n+1)≥(n+1)^3, $ which I wasn't previously aware of. $\endgroup$ Jul 12 '21 at 10:56
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We show that $$ \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n} $$ for any $n \geq 2$.

For any $n \geq 2$ we have that $$ n \int^n_{n-1} \frac{1}{x(x+1)}dx \leq \int^n_{n-1} \frac{1}{x}dx $$ because $\frac{n}{x+1} \leq 1$ for all $x \in [n-1,n]$. But this inequality is equivalent to $$ n \int^n_{n-1} \frac{1}{x}-\frac{1}{x+1}dx \leq \int^n_{n-1} \frac{1}{x}dx \iff\\ \iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^n_{n-1} \frac{1}{x+1}dx \iff\\ \iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^{n+1}_{n} \frac{1}{x}dx \iff\\ \iff (n-1)\ln\left(\frac{n}{n-1}\right) \leq n \ln\left(\frac{n+1}{n}\right) \iff\\ \iff \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}. $$

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We have that

$$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n>0 \iff \ln \left(1+\frac{1}{n+1}\right)^{(n+1)} - \ln \left(1+\frac{1}{n}\right)^n>0$$

and

$$(n+1) \ln \left(1+\frac{1}{n+1}\right) - n \ln \left(1+\frac{1}{n}\right) \ge(n+1)\left[ \ln \left(1+\frac{1}{n+1}\right) - \ln \left(1+\frac{1}{n}\right)\right]=$$

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

$$=(n+1) \ln \left(\frac{(n+2)(n+1)}{(n+1)^2}\right)=(n+1) \ln \left(\frac{n^2+2n+2}{n^2+2n+1}\right) >0$$

since $\log x>0$ for $x>1$.

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Here's a proof I learned by solving problem 90 from Section 11.1 of James Stewart's Calculus: Early Transcendentals (8th edition). We'll need the following result:

Lemma: If $0\leq a<b$ and $n$ is a positive integer, then $$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$$ Proof: Define the function $f:[0,\infty)\rightarrow\mathbb{R}$ by $f(x)=x^{n+1}$, where $n$ is a positive integer. Then $f$ is differentiable (and hence continuous) over $[0,\infty)$. Moreover, by the power rule, $f'(x)=(n+1)x^n$ for all $x\geq 0$, which is strictly increasing because $n$ is a positive integer (this will be important later, so we'll tuck it in our back pocket).
Since $f$ is differentiable over $[0,\infty)$, and $0\leq a<b$, it is necessarily continuous over $[a,b]$ and differentiable over $(a,b)$, so the mean value theorem asserts that there exists $c\in(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a} =\frac{b^{n+1}-a^{n+1}}{b-a}$$ Since $f'$ is strictly increasing and $c<b$, it follows that $f'(c)<f'(b)=(n+1)b^n$, so $$\frac{b^{n+1}-a^{n+1}}{b-a}=f'(c)<f'(b)=(n+1)b^n$$ It immediately follows that $\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$.

Simple algebraic manipulation shows that $\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$ is equivalent to $b^n[(n+1)a-nb]<a^{n+1}$. With this established, we choose $a=1+\frac{1}{n+1}$ and $b=1+\frac{1}{n}$ (this is justified since $\frac{1}{n+1}<\frac{1}{n}$ for all $n>0$, and hence $1+\frac{1}{n+1}<1+\frac{1}{n}$), which gives $$\left(1+\frac{1}{n}\right)^n\left[(n+1)\left(1+\frac{1}{n+1}\right)-n\left(1+\frac{1}{n}\right)\right]<\left(1+\frac{1}{n+1}\right)^{n+1}$$ Therefore, $$\left(1+\frac{1}{n}\right)^n(n+1+1-n-1)<\left(1+\frac{1}{n+1}\right)^{n+1}$$ and thus, $$\left(1+\frac{1}{n}\right)^n(1)=\left(1+\frac{1}{n}\right)^n<\left(1+\frac{1}{n+1}\right)^{n+1}$$ This shows that $U_n=\left(1+\frac{1}{n}\right)^n$ is strictly increasing.

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