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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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    See also http://math.stackexchange.com/questions/64860/proving-bigl1-frac1n1-bigrn1-gt-1-frac1nn2014-03-08

11 Answers 11

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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}. $

39

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}.

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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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    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? ;-)) -Mark2016-08-20
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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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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, 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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    Replace $n$ by $x$ eight times.2014-04-10
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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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    @MarcvanLeeuwen I... would you believe I did this by accident? I think there was another question I had meant to link to, but honestly I can't remember. Thank you for pointing this out.2014-04-08
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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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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}