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I have to show $(1+\frac1n)^n$ is monotonically increasing sequence.

$U_n=(1+\frac1n)^n$, $U_{n+1}=(1+\frac1{n+1})^{n+1}$

I must show $U_{n+1}-U_n\geq0$

i.e. to show $(1+\frac1{n+1})^{n+1}-(1+\frac1n)^n\geq0$

I am trying to go ahead of this step.

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    Unless I'm missing something, that sequence is increasing, not decreasing.2012-07-07
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    @Javier Badia :-May be you are correct. But I am told it is decreasing. Let us see if we can prove it is increasing2012-07-07
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    Well, if you wish to prove something which is provably false you might as well begin by assuming $0=1$... You should contact whoever told you it is decreasing and correct them.2012-07-07
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    @Asaf: rajendra's comment says "Let us see if we can prove it is *increasing*."2012-07-07
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    @anon: But the question doesn't...2012-07-07
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    I thought you were responding to the comment, sorry.2012-07-07
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    Since $U_1-U_2=-\frac14<0$ it is obviously impossible to prove what was originally asked in the question (prove a decreasing sequence). I've taken the freedom to modify the question so that it does not ask for the impossible.2012-07-07
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    Thanks Marc van Leeuwen. I should have verified it earlier.2012-07-07
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    If you have $(1+\frac{1}{n})^{n+1}$ then decreasing is a right fact.2012-07-07
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    th sequence is monotone increasing2014-03-08
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    See also http://math.stackexchange.com/questions/64860/proving-bigl1-frac1n1-bigrn1-gt-1-frac1nn2014-03-08

10 Answers 10

2

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 $t_n < t_{n+1}$.

35

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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    @Olivier: Thanks for editing.2012-07-07
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    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.2012-07-07
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    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. Anyway, this proof is more elementary. +12012-07-08
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    @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. Thanks2012-07-08
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    @Babak Sorough Thanks. I have chosen this answer. $$Will you please explain$$ $$How\bigg(1−\frac{n}{(n+1)^2}\bigg)\geq\bigg(\frac{1}{1+\frac{1}{n+1}}\bigg)?$$2012-07-11
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    @rajendrabakre: Thanks for choosing mine, but note that other answers like Olivier's will give you more insight about the question. I edited the answer. :)2012-07-11
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    ++Helpful, once again! :^D2013-03-08
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    @amWhy: Thanks Amy ;-)2013-03-08
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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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    i tried its too complicated to conclusively tell it is always positive2014-03-08
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    @raj okay I am adding the differential in.2014-03-08
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    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?2014-03-08
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    @digital-Ink that depends on your definition of Euler's constant2014-03-08
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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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    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?2014-03-08
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    @digital-Ink I define $\ln x$ to be the function whose derivative is $1/x$ and whose value at $x = 1$ is $0$.2014-03-08
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    @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.2014-03-08
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    @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.2014-03-08
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    @omnomnomnom how can you say $\ln\left(1 + \frac 1x\right) \geq \frac{1}{x+1}$2014-03-08
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    It's a little easier to deal with in the form I have it in now2014-03-08
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    @Sabyasachi: I remember well that in high school we used just algebraic properties (like the binomial expansion formula) to prove that $a_n$ is increasing and bounded (actually, we used it together with $b_n=(1+\frac{1}{n})^{n+1})$, then we introduced $\mathrm{e}$ as the limit of these sequences. This seems the most elementary.2014-03-08
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    @Sabyasachi: Clearly, the path to the truth are many. But it seems wrong to go around through all the Calculus (derivatives, anti-derivatives, integrals or series), just to prove a fundamental property. At least, that's my view.2014-03-08
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    @digital-Ink yes "elementary" yes. fun? no.2014-03-08
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    @digital-Ink clearly yes and I agree. but for some reason I view calculus as completely fundamental, as fundamental as arithmetic. Yes, it took us a lot of time to *develop* calculus, but once you have understood it, isn't the beauty immediately apparent? Binomial theorem will get messy here, although in some cases that IS applied elegantly, it might be better than the application of calculus here. Feel free to post something like that if you have it. Cheers. :)2014-03-08
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    @Sabyasachi: I will try to write an algebraic proof. Unless someone else writes it first.2014-03-08
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    @ Omnomnomnom we can take $ g(x) = \ln(x+1) - (\ln(x) + \frac{1}{x+1} ) $ and prove that g(x) is increasing2014-03-08
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    @raj I suppose so; all we really need to show is that it is always positive. See the link I've provided at the end of my answer for an alternate approach.2014-03-08
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    @digital-Ink that'd be great. variety is the spice of life. :D2014-03-08
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    @Sabyasachi: Done. No need for binomial expansion.2014-03-08
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    @Sabyasachi: Also, I assure you that there is no need for calculus in order to prove the inequality between the geometric mean and the arithmetic mean (of course, using the logarithm again as a concave function trough Jensen's inequality would be elegant)2014-03-08
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    @digital-Ink yes I am aware of the fact that there is no need of calculus to prove $AM \ge GM$. I lost you though, where does this come in here?2014-03-08
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    oh. point in sarcasm? Got it.2014-03-08
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    @Sabyasachi: I used the inequality in order to prove that $a_n$ is increasing. Please, see below. No sarcasm included! Just wanted to point out that the inequality between means can be considered an algebraic tool.2014-03-08
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    Then link at the end of this answer is one to the very _question_ above. Is this some kind of humour that I am missing?2014-04-08
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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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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 $$ n \int^n_{n-1} \frac{1}{x}-\frac{1}{x+1}dx \leq \int^n_{n-1} \frac{1}{x}dx \iff\\ \iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^n_{n-1} \frac{1}{x+1}dx \iff\\ \iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^{n+1}_{n} \frac{1}{x}dx \iff\\ \iff (n-1)\ln\left(\frac{n}{n-1}\right) \leq n \ln\left(\frac{n+1}{n}\right) \iff\\ \iff \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}. $$