Question

Show that the inequality \dfrac{6^n}{n!}\leq \dfrac{6^5}{5!}\times \dfrac{6}{n} is true

Collected in the board: Inequality

zheng xs posted 1 year ago

Answer

Let's evaluate the expression \dfrac{6^n}{n!}\cdotp n

When n=6

\dfrac{6^6}{6!}\cdotp 6= \dfrac{6^5}{5!}\times\dfrac{6}{6}\times 6 = \dfrac{6^5}{5!}\times 6

When n=7

\dfrac{6^7}{7!}\cdotp 7= \dfrac{6^5}{5!}\times\dfrac{6^2}{6\times 7 }\times 7 = \dfrac{6^5}{5!}\times 6

When n=8

\dfrac{6^8}{8!}\cdotp 8= \dfrac{6^5}{5!}\times\dfrac{6^3}{6\times 7\times 8 }\times 8 < \dfrac{6^5}{5!}\times 6

\dots

\dfrac{6^n}{n!}\cdotp n = \dfrac{6^5}{5!}\times \dfrac{6^{n-5}}{6\times 7\dots n }\cdotp n =\dfrac{6^5}{5!}\times \dfrac{6^{n-5}}{6\times 7\dots n-1 }

=\dfrac{6^5}{5!}\times \dfrac{6^{n-6}}{6\times 7\dots n-1 } \times 6

< \dfrac{6^5}{5!}\times 1 \times 6

In summary, we have verified that the inequality \dfrac{6^n}{n!}\leq \dfrac{6^5}{5!}\times \dfrac{6}{n} is true for n\geq 6


Steven Zheng posted 1 year ago

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