Using the Cauchy-Schwarz Inequality, we have $$ \begin{align} 1 &=\left(\int_n^{n+1}1\,\mathrm{d}x\right)^2\\ &\le\left(\int_n^{n+1}x\,\mathrm{d}x\right)\left(\int_n^{n+1}\frac1x\,\mathrm{d}x\right)\\ &=\left(n+\frac12\right)\log\left(1+\frac1n\right) \end{align} $$ which means that $$ \left(1+\frac1n\right)^{n+1/2}\ge e $$ This hints that $\left(1+\frac1n\right)^{n+1/2}$ might be decreasing.

In this answer, it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. The proofs use Bernoulli's Inequality. However, applying Bernoulli to $\left(1+\frac1n\right)^{n+1/2}$ is inconclusive.

Attempt to show decrease: $$ \begin{align} \frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}} &=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\ &\ge\left(1+\frac{2n}{n^2-1}\right)\frac{n-1}{n+1}\\[6pt] &=1-\frac{2}{(n+1)^2} \end{align} $$ Attempt to show increase: $$ \begin{align} \frac{\left(1+\frac1n\right)^{2n+1}}{\left(1+\frac1{n-1}\right)^{2n-1}} &=\left(1-\frac1{n^2}\right)^{2n}\frac{n+1}{n-1}\\ &\ge\left(1-\frac2n\right)\frac{n+1}{n-1}\\[6pt] &=1-\frac{2}{n(n-1)} \end{align} $$ Neither works.

Without resorting to derivatives, is there something stronger than Bernoulli, but similarly elementary, that might be used to show that $\left(1+\frac1n\right)^{n+1/2}$ decreases?

  • $\begingroup$ A numerical test gives $f(5577) - f(5578) \sim -5\cdot 10^{-13}$, and several others. I'm not even sure if this is a numerical error, as it is not that small. $\endgroup$ Jun 12, 2019 at 4:32
  • $\begingroup$ Well, that's pretty much a numerical error, since wolfram alpha yields much smaller estimate. In any case, a solution if exists, must be very fine-tuned (like yours below). $\endgroup$ Jun 12, 2019 at 4:35
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    $\begingroup$ @QuangHoang. It seems that you used $n$ instead of $\frac 1n$. Computing $f(5577) - f(5578) $ gives $2.61 \times 10^{-12}$. $\endgroup$ Jun 12, 2019 at 4:57
  • $\begingroup$ It turns out that using the case $m=2$ of Theorem $1$ from my answer below, one gets $$ 1-\frac1{3n(n-1)^2} \le\left(1-\frac1{n^2}\right)^{2n}\frac{n+1}{n-1} \le1-\frac1{3n(n+1)^2} $$ $\endgroup$
    – robjohn
    Jun 12, 2019 at 16:21

4 Answers 4


Preliminaries: A couple of extensions to Bernoulli's Inequality.

Bernoulli's Inequality says that $(1+x)^n$ is at least as big as the first two terms of its binomial expansion. It turns out, at least for $n\in\mathbb{Z}$, that a sharper inequality can be obtained using any partial sum with an even number of terma.

Theorem $\bf{1}$: for $m\ge1$, $n\ge0$, and $x\gt-1$, $$ (1+x)^n\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag1 $$

Proof (Induction on $n$): $(1)$ is trivial for $n=0$. Assume $(1)$ is true for $n-1$, then $$ \begin{align} (1+x)^n &=(1+x)(1+x)^{n-1}\tag{1a}\\[9pt] &\ge(1+x)\sum_{k=0}^{2m-1}\binom{n-1}{k}x^k\tag{1b}\\ &=\sum_{k=0}^{2m-1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]x^k+\binom{n-1}{2m-1}x^{2m}\tag{1c}\\ &\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: factor
$\text{(1b)}$: assumption for $n-1$
$\text{(1c)}$: multiply sum by $1+x$
$\text{(1d)}$: Pascal's Rule

Thus, $(1)$ is true for $n$.

Theorem $\bf{2}$: for $m\ge1$, $n\ge0$, and $x\gt-1$, $$ (1+x)^{-n}\ge\sum_{k=0}^{2m-1}\binom{-n}{k}x^k\tag2 $$

Proof (Induction on $n$): Note that another way of writing $(2)$ is $$ (1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\le1\tag{2a} $$

$\text{(2a)}$ is trivial for $n=0$. Assume $\text{(2a)}$ is true for $n-1$, then $$ \begin{align} &(1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\\ &=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k(1+x)\tag{2b}\\ &=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k{\textstyle\left[\binom{n+k-1}{k}-\binom{n+k-2}{k-1}\right]}x^k-{\textstyle\binom{n+2m-2}{2m-1}}x^{2m}(1+x)^{n-1}\tag{2c}\\ &=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-2}{k}x^k-\binom{n+2m-2}{2m-1}x^{2m}(1+x)^{n-1}\tag{2d}\\[9pt] &\le1\tag{2e} \end{align} $$ Explanation:
$\text{(2b)}$: factor
$\text{(2c)}$: multiply sum by $1+x$
$\text{(2d)}$: Pascal's Rule
$\text{(2e)}$: assumption for $n-1$

Thus, $\text{(2a)}$ is true for $n$.

Note that for positive integer exponents, Bernoulli's Inequality is the case $m=1$ of Theorem $1$, and for negative integer exponents, it is the case $m=1$ of Theorem $2$.

Answer: Use the case $m=2$ of Theorem $1$: $$ \begin{align} &\frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}}\\ &=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\ &\ge\left(1+\frac{2n}{n^2-1}+\frac{2n(2n-1)}{2\left(n^2-1\right)^2}+\frac{2n(2n-1)(2n-2)}{6\left(n^2-1\right)^3}\right)\frac{n-1}{n+1}\\ &=1+\frac{n^2+n+6}{3(n-1)(n+1)^4}\\[9pt] &\ge1 \end{align} $$ That is, $\left(1+\frac1n\right)^{n+1/2}$ is decreasing.


Another simple way is you can derive: $$\dfrac{1}{2}\ln\dfrac{1+x}{1-x} = x + \dfrac{x^3}{3} + \dfrac{x^5}{5} \dots$$ by only using Taylor's expansion and then substitute $x = \dfrac{1}{2n+1}:$ $$\left(n+\frac 12\right)\ln\left(1+\frac 1n\right) = 1 + \dfrac{1}{3(2n+1)^2} + \dfrac{1}{5(2n+1)^4}+\dots$$ and it's clear that the above is decreasing.

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    $\begingroup$ Amazing! This should be the best solution. $\endgroup$
    – orangeskid
    Jul 5 at 23:50
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    $\begingroup$ Nice. Typos: it should be $1 + \frac{1}{3(2n+1)^2} + \frac{1}{5(2n+1)^4} + \cdots$ $\endgroup$
    – River Li
    Jul 6 at 0:38
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    $\begingroup$ @orangeskid I thought so but then I later realized OP asked for ways without derivatives and that makes me think OP's own inductive solution is perhaps the best one given the requirement. But yes, I do think this particular form of Taylor expansion of $\ln\frac{1+x}{1-x}$ is very useful to know for beginning analysis students. I sure had my minds blown away the first time I saw these kinds of things. $\endgroup$
    – dezdichado
    Jul 6 at 6:03
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    $\begingroup$ I used a similar argument in connection with the Stirling formula some time ago. $\endgroup$
    – Gary
    Jul 6 at 6:08

We'll prove that this sequence indeed decreases.

We need to prove that $$\left(1+\frac{1}{n+1}\right)^{n+\frac{3}{2}}<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}.$$Let $$f(x)=\left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)-\left(x+\frac{3}{2}\right)\ln\left(1+\frac{1}{1+x}\right)=$$ $$=(2x+2)\ln(1+x)-\left(x+\frac{1}{2}\right)\ln{x}-\left(x+\frac{3}{2}\right)\ln(x+2),$$ where $x>0$.

But $$f''(x)=\frac{2}{x^2(x+2)^2(x+1)}>0$$ and $$\lim_{x\rightarrow+\infty}f'(x)=\lim_{x\rightarrow+\infty}\left(\ln\frac{(x+1)^2}{x(x+2)}+2-\frac{x+\frac{1}{2}}{x}-\frac{x+\frac{3}{2}}{x+2}\right)=0,$$ which says that $f'(x)<0$, $f$ decreases and since it's obvious that $$\lim_{x\rightarrow+\infty}f(x)=0,$$ we obtain that $f(x)>0$ and our sequence indeed decreases.

  • $\begingroup$ It looks as if you want to prove that $f(x)\ge0$, not $f'(x)\le0$. In any case, Bernoulli is often used before derivatives are introduced because it has a simple inductive proof. In the question, integration was used to derive that $\left(1+\frac1n\right)^{n+1/2}\ge e$, but if the question has an elementary answer, it also serves as an elementary proof that $\left(1+\frac1n\right)^{n+1/2}\ge e$. Proofs avoiding power series were explicitly requested, but perhaps avoidance of derivatives should be explicitly requested. $\endgroup$
    – robjohn
    Jun 12, 2019 at 3:21

The function $$(0, \infty) \ni x \mapsto f(x) \colon =\log(1+\frac{1}{x}) \cdot (x + \frac{1}{2})-1$$ has second derivative $$f''(x) = \frac{1}{2 x^2(x+1)^2} > 0$$ is positive, and has limit $0$ at $\infty$, so it has to be decreasing( hand-waving here). If we look at higher order derivatives we notice that their sign alternate. That gets us thinking that $f$ is totally monotone. It is, in fact - one checks that $f(x)$ is the Laplace transform of the positive function $$g(t) = \frac{2+t + e^t(-2 + t)}{2e^t t^2}$$ (the numerator has a positive series expansion starting with $t^3/6$)

$\bf{Added:}$ We can try other tricks with the Laplace transform. If we take the Laplace transform of the positive function $$h(t) = \frac{-12 - 6 t - t^2 + e^t( 12 - 6 t + t^2)}{2 t^4}$$ we get

$$k(x) = \frac{1}{12} - x(x+1) [\, \log ( 1+ \frac{1}{x}) (x+\frac{1}{2})-1 \,]$$

so further info on $f(x)$.

Note: the functions in $t$ come from Pade approximants for $e^{-t}$.

$\bf{Added:}$ Another solution that seems more elementary.

To show that $\log(1+\frac{1}{x}) \cdot (x + \frac{1}{2})$ is decreasing it is enough to show that it is the limit of a sequence of decreasing functions.

Now, given $a_1$, $\ldots$, $a_N$ positive numbers the function

$$(0, \infty) \ni x \mapsto ((a_1+x) + \cdots + (a_N + x))\left( \frac{1}{a_1 + x} + \cdots + \frac{1}{a_N + x} \right)$$ is decreasing ( the proof is left for another problem).


$$\lim_{N\to \infty}\frac{1}{N^2} ( (x+\frac{1}{N} ) + \cdots + (x+\frac{N}{N}) ) \left( \frac{1}{x+\frac{1}{N}} + \cdots + \frac{1}{x+\frac{N}{N}} \right)$$

is decreasing. But the limit is $(x+\frac{1}{2}) \log (1+ \frac{1}{x})$.

Note: this shows the function is decreasing, but perhaps not strict. However, we can estimate the integral with lower and upper sums and get the function is strict decreasing.


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