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I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.

$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)

I was led to this problem by trying to prove by induction that $\big(1+\frac{1}{n}\big)^n\le3-\frac{1}{n}$, as in

$\hspace{.4 in}$ A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?

Community
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user84413
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    Have you tried showing the analogous statement for the function $x\in(0,\infty)\mapsto (1+\frac{1}{x})^x+\frac{1}{x}$? This may be much easier (you can differentiate and rely on every real analysis tool you have), and would imply what you want. – Clement C. Aug 28 '15 at 22:50

4 Answers4

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Let $$ \eqalign{f(n) = \dfrac{1}{n} + \left( 1 + \dfrac{1}{n}\right)^n &= \dfrac{1}{n} + \exp\left( n \ln\left(1+\dfrac{1}{n}\right)\right) \cr &= \dfrac{1}{n} + \exp\left(1 - \dfrac{1}{2n} + \dfrac{1}{3n^2} + O\left(\dfrac{1}{n^3}\right)\right) \cr &= e - \dfrac{e-2}{2n} + \dfrac{11e}{24 n^2} + O\left(\dfrac{1}{n^3}\right) }$$

Then $$\eqalign{f(n+1) &= e - \dfrac{e-2}{2n+2} + \dfrac{11e}{24 (n+1)^2} + O\left(\dfrac{1}{n^3}\right)\cr &= e - \dfrac{e-2}{2n} + \dfrac{23 e - 24}{24 n^2} + O\left(\dfrac{1}{n^3}\right) \cr f(n+1) - f(n) &= \dfrac{12e-24}{24n^2} + O\left(\dfrac{1}{n^3}\right)}$$

and since $e > 2$, this is positive for sufficiently large $n$.

Robert Israel
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As suggested by Clement C., let: $$ f(x)=\left(1+\frac{1}{x}\right)^{x}+\frac{1}{x}.\tag{1}$$ Then: $$ f'(x) = \left(1+\frac{1}{x}\right)^{x}\left(\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right)-\frac{1}{x^2}\tag{2} $$ but, due to convexity: $$\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}=-\frac{1}{x+1}+\int_{x}^{x+1}\frac{dt}{t}\geq \frac{1}{2(x+1)^2}\tag{3}$$ hence for any $x\geq 8$: $$ f'(x)\geq \frac{\left(1+\frac{1}{8}\right)^8}{2(x+1)^2}-\frac{1}{x^2}>0.\tag{4} $$

Jack D'Aurizio
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  • How did you find the "$8$?" We can find it numerically. But I was curious if there were an analytical method you used here. If not, then why not save a step and find numerically that value of $x$ for which $f'(x)$ equals zero? That value is roughly $5.474$, I believe. – Mark Viola Aug 29 '15 at 05:04
  • @Dr.MV: no magic this time, $8$ is just a number for which the chain of inequalities is true. I tried $4$ before: $\left(1+\frac{1}{4}\right)^4$ is well above $2$, but $\frac{(1+1/4)^4}{2(x+1)^2}-\frac{1}{x^2}>0$ holds only for big $x$. With $8$ everything goes smooth, just by trial and error. – Jack D'Aurizio Aug 29 '15 at 05:49
  • Yes, I suspected. But then curious as to why not just save a step? – Mark Viola Aug 29 '15 at 06:01
  • @Dr.MV: how to save a step? If we take $x\geq 8$, both $\left(1+\frac{1}{x}\right)^x\gg 2$ and $\frac{\left(1+\frac{1}{x}\right)^x}{2(x+1)^2}-\frac{1}{x^2}>0$ hold, hence we actually save a step. – Jack D'Aurizio Aug 29 '15 at 13:32
  • See my previous comment. Rather than finding a bound for $\log(1+1/x)-1/(x+1)$ and then finding that $x>8$ renders $f'$ positive, simply find that $x>6$ renders $f'$ positive. – Mark Viola Aug 29 '15 at 15:11
  • @Dr.MV: even if you find numerically the value for which $f^\prime$ cancels, you still have to show that $f^\prime$ itself doesn't calcel afterwards (or becomes negative) -- don't you? – Clement C. Aug 29 '15 at 15:18
  • @clementc. Yes, I see. Since all the terms decline as $1/x$, there wouldn't be anymore turning points after $f'=0$, right? – Mark Viola Aug 29 '15 at 15:26
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Let $a_n = (1 + 1/n)^n.$

We want to show $a_{n+1} - a_{n} \geq \dfrac{1}{n(n+1)}$ for large $n$.

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

The RHS can be expanded as

$\left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1} = \dfrac{n+1}{n} \times \left( \underbrace{1 - \dfrac{1}{n+1}} + \dfrac{1}{2!(n+1)^2}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} - \dfrac{1}{3!(n+1)^3}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} \left(1 - \dfrac{2}{n+1} \right) + \dots (-1)^{n+1} \dfrac{1}{(n+1)!(n+1)^{n+1}}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} \left(1 - \dfrac{2}{n+1} \right) \dots\left(1 - \dfrac{n}{n+1}\right)\right).$

Since $ \dfrac{n+1}{n} \times (1-\dfrac{1}{n+1}) = 1$, we have $\dfrac{a_{n+1}}{a_n} = 1 + \dfrac{1}{2!(n+1)^2} - \dfrac{1}{3!(n+1)^3} \left(1 - \dfrac{2}{n+1} \right) + \dots$

So,

$|\dfrac{a_{n+1}}{a_n} - 1 - \dfrac{1}{2(n+1)^2}| \leq \dfrac{1}{3!(n+1)^3} + \dfrac{1}{4!(n+1)^4} + \dots \leq \dfrac{1}{6(n+1)^2n}.$

This implies $$ (n+1)^2 \left( \dfrac{a_{n+1}}{a_n} - 1 \right) \to 1/2$$ so upon multiplying the above by $na_n/(n+1)$ $$ n(n+1)(a_{n+1} - a_n) \to e/2 > 1.$$ Hence, $a_{n+1} - a_n \geq 1/n(n+1)$ for all large n.

Arin Chaudhuri
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  • Could you please explain how you expanded the RHS; I think it's correct, but I don't see how you got it in this form. Also, would you please explain how you got the term $\frac{1}{4!(n+1)^4}$. – user84413 Aug 29 '15 at 23:02
  • Use the Binomial theorem, the $k^{th}$ term is \begin{align} (-1)^k {n+1 \choose k} \frac{1}{(n+1)^{2k}} &=\frac{(-1)^k}{k!} \frac{1}{(n+1)^k} \frac{ (n+1)(n+1 - 1) \dots (n+1 - (k-1))}{(n+1)(n+1) \dots (n+1)} \ &= \frac{(-1)^k}{k!} \dfrac{1}{(n+1)^k} \left(1 - \frac{1}{n+1}\right) \dots \left(1 - \frac{k-1}{n+1}\right). \end{align}

    The term you are interested comes from setting $k = 4$

     – Arin Chaudhuri Aug 30 '15 at 00:43
  • Also note that the terms of the form $1 - \dfrac{k}{n+1}$ are less than 1 in absolute value. – Arin Chaudhuri Aug 30 '15 at 00:55
  • Thanks -- this is a nice argument. – user84413 Aug 30 '15 at 01:08
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Here is another way to approach this problem. The function $$f(z) = 1 - z/2 + z^2/3 + \ldots + (-1)^{k+1} z^k/(k+1) + \ldots $$ is analytic on the unit disc $ \{ z : |z| < 1\}$, which implies $ g(z) = \exp f(z)$ is also analytic on $ \{ z : |z| < 1\}$ and hence can be expanded as a power series $$g(z) = a_0 + a_1 z + a_2 z^2 + \dots + $$ in $ \{ z : |z| < 1\}$. We can easily compute the the first few coefficients as $a_0 = \exp f(0) = e$, $a_1 = \exp f(0) f^{'}(0) = -e/2$, and similarly $a_3 = 11e/24.$

However, $f(x) = \log(1+x)/x$ for for all real $x$ with $ 0 < x < 1$, so $g(x) = (1+x)^{1/x}$ for $ 0 < x < 1$ and the above series for real $x$ is an analytic extension of $(1+x)^{1/x}$ to $-1 < x < 1$.

Writing $$(1+x)^{1/x} = e - ex/2 + 11x^2/24 + \dots $$ from which we get $(1+x)^{1/x} + cx = e + (c - e/2) x + 11ex^2/24 + \dots + \dots $.

The derivative of the above function at 0 is c -e/2, which is < 0, if c < e/2, by the continuity of the derivative, there is an interval $[0,\epsilon]$ on which the derivative of the function above is strictly negative and hence it decreases. Since 1/n decreases and lies in $[0,\epsilon]$ for all large $n$ this means $(1+1/n)^{n} + c/n$ increases eventually for any $ c < e/2$. This holds for any $x_n$ that strictly decreases to 0 not only $1/n,$, $(1+x_n)^{1/x_n} + c x_n$ eventually increases if $ c < e/2$. We can similarly argue that $(1+x_n)^{1/x_n} + c x_n$ eventually increases if $ c > e/2$ if $x_n$ strictly decreases to 0. For $c = e/2$, the positivity of the coefficient of $x^2$ implies $(1+x_n)^{1/x_n} + e x_n / 2$ eventually starts decreasing.

Arin Chaudhuri
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