Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing

Question

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.

Answer 1

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 $a_{n}<a_{n+1}$.

Answer 2

Let $x_n=\big(1+\frac{1}{n}\big)^n$ be the $n^\text{th}$ element of the sequence. Then $$ \begin{align} \frac{x_{n+1}}{x_{n}} &=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}} \\&=\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg) \\&=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg) \\&≥^*\bigg(1-\frac{n}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg) \\&>\bigg(1-\frac{n+1}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg) \\&=1 \end{align}$$ The $\ge^*$ step follows from Bernoulli's inequality, which says that $(1+x)^m\ge 1+mx $ whenever $m\in \mathbb N$ and $x\ge -1$.

Answer 3

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.

Answer 4

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 $e$.)

Answer 5

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.

Answer 6

My two cents (as later similar question closed):

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

Answer 7

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 $t_n < t_{n+1}$.

Answer 8

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

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

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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.

Answer 9

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 $x = 2014$.

Answer 10

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}

Answer 11

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 $$ n \int^n_{n-1} \frac{1}{x}-\frac{1}{x+1}dx \leq \int^n_{n-1} \frac{1}{x}dx \iff\\ \iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^n_{n-1} \frac{1}{x+1}dx \iff\\ \iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^{n+1}_{n} \frac{1}{x}dx \iff\\ \iff (n-1)\ln\left(\frac{n}{n-1}\right) \leq n \ln\left(\frac{n+1}{n}\right) \iff\\ \iff \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}. $$

Answer 12

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 $x>1$.

Answer 13

By Lagrange's Mean Value Theorem.

$$ \exists\xi\in(x,x+1):\frac{\ln(x+1)-\ln x}{(x+1)-x}=\frac{1}{\xi}>\frac{1}{x+1} $$

Thus,

$$ \frac{d}{dx}\ln\left(1+\frac1x\right)^x=\ln(x+1)-\ln x-\frac{1}{x+1}>0 $$

This shows that $\left(1+\dfrac1x\right)^x$ is monotonically increasing.

Answer 14

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.