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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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11 Answers 11

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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. $\endgroup$ – Olivier Bégassat 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 $\endgroup$ – mrs Jul 8 '12 at 5:10
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    $\begingroup$ @Babak Sorough Thanks. I have chosen this answer. $$Will you please explain$$ $$How\bigg(1−\frac{n}{(n+1)^2}\bigg)\geq\bigg(\frac{1}{1+\frac{1}{n+1}}\bigg)?$$ $\endgroup$ – rajendra bakre 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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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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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 $n<k$, 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? $\endgroup$ – digital-Ink Mar 8 '14 at 19:11
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    $\begingroup$ @digital-Ink that depends on your definition of Euler's constant $\endgroup$ – Omnomnomnom 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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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? $\endgroup$ – digital-Ink 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 $0$. $\endgroup$ – Omnomnomnom 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
  • $\begingroup$ @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? $\endgroup$ – Aditya Dutt Jul 26 '19 at 17:26
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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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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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