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Prove that the sequence $a_n=\frac{n+1}{n!}$ is decreasing after the second term.

I thought that if $a_{n+1}<a_n$ then it is decreasing

After some work , $$\frac{n+2}{n+1}<n+1$$

Then im not sure where to go in order to prove this .

Nick Peterson
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cindy
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2 Answers2

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Hint: $$\frac{n+1}{n!} = \frac{1}{(n-1)!}+\frac{1}{n!}.$$

njguliyev
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Calculate $\dfrac{a_{n+1}}{a_n}$. We have $$\frac{a_{n+1}}{a_n}=\frac{(n+2)/(n+1)!}{(n+1)/n!}=\frac{n+2}{(n+1)^2}.$$

We want to show that $\dfrac{n+2}{(n+1)^2}\lt 1$ if $n\ge 1$. Equivalently, we want to show that $(n+1)^2\gt n+2$. Since $(n+1)^2=n(n+2)+1$, it is clear that $(n+1)^2\gt n+2$ if $n\ge 1$.

André Nicolas
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  • can you tell me the reason for calculating $\dfrac{a_{n+1}}{a_n}$. – cindy Oct 02 '13 at 18:04
  • We want to show that $a_{n+1}\lt a_n$. (That's what decreasing means.) Since our sequence consists of positive numbers, the inequality $a_{n+1}\lt a_n$ is equivalent to $\frac{a_{n+1}}{a_n}\lt 1$ (we divided both sides by $a_n$). In general, there are two sorts of common strategies for showing that $x\lt y$. (i) if $y$ is positive, show that $\frac{x}{y}\lt 1$; (ii) show that $y-x\gt 0$ (that would work nicely here too). – André Nicolas Oct 02 '13 at 18:11