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I'm looking for a proof of $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$, optionally $\ln(n)+\frac{1}{n+1}\leq \ln(n+1)$

Matthew
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    Let $f(x) = \ln(1 +x) - \frac x{x+1}$. $f'(x) \geq 0$ if $x \geq 0$ then $f(x) \geq f(0) =0$ for any $x\geq 0$. Taking $x = \frac1n$. – nguyen0610 Jan 27 '16 at 00:25

5 Answers5

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If $x \ge 0$,

$$e^x = 1 + \int_0^x e^t\, dt \le 1 + \int_0^x e^x\, dt = 1 + xe^x.$$

So $(1 - x)e^x \le 1$, and thus

$$e^x \le \frac{1}{1 - x},\quad 0 \le x < 1.$$

Letting $x = \frac{1}{n+1}$, we obtain

$$e^{\frac{1}{n+1}} \le \frac{1}{1 - \frac{1}{n+1}} = \frac{n+1}{(n+1)-1} = \frac{n+1}{n} = 1 + \frac{1}{n}.$$

Equivalently,

$$e^{\frac{1}{n+1}} - 1 - \frac{1}{n} \le 0.$$

kobe
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Since $\ln(n+1)-\ln n$ represents the area under $y=\frac{1}{x}$ from $n$ to $n+1$ and $y=\frac{1}{x}$ is decreasing,

we get that $\ln(n+1)-\ln n>\frac{1}{n+1}$.

Alternatively, applying the Mean Value Theorem to $f(x)=\ln x$ on $[n,n+1]$ gives

$\ln(n+1)-\ln n=f(n+1)-f(n)=f^{\prime}(c)=\frac{1}{c}>\frac{1}{n+1}$ for some $c$ in $(n,n+1)$.

user84413
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In THIS ANSWER and THIS ONE, I showed using only (i) the limit definition of the exponential function and (ii) Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\frac{1}{1-x}\ge e^x\ge 1+x \tag 1$$

for $x<1$. Letting $x=1/(n+1)$ in the left-hand inequality in $(1)$ yields

$$e^{1/(n+1)}\le \frac{1}{1-\frac{1}{n+1}}=1+\frac1n \tag 2$$

Rearranging $(2)$, we obtain the desired inequality

$$\bbox[5px,border:2px solid #C0A000]{e^{1/(n+1)}-1-\frac1n\le 0}$$

And we are done!

for

Community
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Mark Viola
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for $0 < x < 1$, $e^x =\sum_{n=0}^{\infty} \frac{x^n}{n!} \lt\sum_{n=0}^{\infty} x^n =\frac1{1-x} $.

Therefore $e^{1/(n+1)} <\frac1{1-1/(n+1)} =\frac{n+1}{n} =1+\frac1{n} $.

marty cohen
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In this answer, it is shown that $$ \left(1+\frac1n\right)^{n+1} $$ is a decreasing sequence. Since $e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{n+1}$, we have $$ e\lt\left(1+\frac1n\right)^{n+1} $$ Taking $n+1^{\text{st}}$ roots and subtracting yields $$ e^{\large\frac1{n+1}}-1-\frac1n\lt0 $$

Community
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robjohn
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