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Can anyone explain why,

$$\displaystyle 1 - e^{-\frac{1}{k}} \geq \frac{1}{ke}$$

for $k \geq 1$.

This inequality is used to show that the series

$$\sum_{k=1}^\infty 1 - e^{-\frac{1}{k}} $$

is divergent, and I do not see how one can derive the result.

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3 Answers

up vote 2 down vote accepted

Lets use only our knowledge of $e$, and a geometric series identity. Notice that $$\left(1-e^{-\frac{1}{k}}\right)\left(1+e^{-\frac{1}{k}}+e^{-\frac{2}{k}}\cdots+e^{-\frac{k-1}{k}}\right)=1-e^{-1}.$$ Now, $e^{-x}\leq1$ for $x\geq0,$ so $$\left(1-e^{-\frac{1}{k}}\right)k=\left(1-e^{-\frac{1}{k}}\right)\left(1+1+1\cdots+1\right)$$

$$\geq\left(1-e^{-\frac{1}{k}}\right)\left(1+e^{-\frac{1}{k}}+e^{-\frac{2}{k}}\cdots+e^{-\frac{k-1}{k}}\right) $$

$$=1-e^{-1}$$ $$\geq\frac{1}{e},$$ and so $$1-e^{-\frac{1}{k}}\geq\frac{1}{ke}.$$

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You actually have equality in one step there, although equality does imply $\geq$... –  Thomas Andrews Dec 18 '12 at 0:31
    
@ThomasAndrews: Thanks, fixed. –  Eric Naslund Dec 18 '12 at 0:33
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$$e^x \left(1 - \dfrac{x}{e}\right) \geq 1$$ for $x \leq 1$.

Consider $f(x) = e^x - x e^{x-1}$. We then have $$f'(x) = e^x - x e^{x-1} - e^{x-1} = e^{x-1} \left(e-x-1\right) > 0$$ for $x \leq 1$. Hence, $f(x)$ is an increasing function for $x \leq 1$. Hence, $$f(x) \geq f(0) \implies e^x \left(1 - \dfrac{x}{e}\right) \geq 1$$ Choosing $x = \dfrac1k$, we get $$e^{1/k} \left(1 - \dfrac1{ke}\right) \geq 1 \implies e^{-1/k} \leq 1 - \dfrac1{ke} \implies 1 - e^{-1/k} \geq \dfrac1{ke}$$

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Since $\frac{k+1}{k}=1+\frac{1}{k}\leq e^{\frac{1}{k}}$, therefore $e^{-1/k}\leq \frac{k}{k+1}$. Thus, $\frac{1}{k+1}\leq 1-e^{-1/k}$ . This inequality will do the job.

However, if you still want to show your inequality you can use the fact that $\forall k\geq 1[\frac{1}{k+1}\geq \frac{1}{ke}]$

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