Can someone please help me figure out why it is enough to take $n\geq 9 $ in order to get that: $\displaystyle \frac{ 9 \cdot 2^{n+1} } {(n+1)!} < \frac{1}{100}\;\;?$
Thanks in advance
Can someone please help me figure out why it is enough to take $n\geq 9 $ in order to get that: $\displaystyle \frac{ 9 \cdot 2^{n+1} } {(n+1)!} < \frac{1}{100}\;\;?$
Thanks in advance
First note that $a_n = \dfrac{2^{n+1} \cdot 9}{(n+1)!}$ is a monotone decreasing sequence for $n \geq 1$. Hence, if $a_n < \epsilon$, then $a_m < \epsilon$ for all $m \geq n$. $a_n = \dfrac{2^{n+1} \cdot 9}{(n+1)!} = \dfrac{3 \cdot 2^{n-2} \cdot 4!}{(n+1)!}$ $a_8 = \dfrac{3 \cdot 2^6 \cdot 4!}{9!} = \dfrac{3 \cdot 2^6}{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5} = \dfrac{2^2}{9 \cdot 7 \cdot 5} = \dfrac4{315} > \dfrac4{400} = \dfrac1{100}$ $a_9 = \dfrac{3 \cdot 2^7 \cdot 4!}{10!} = \dfrac{3 \cdot 2^7}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5} = \dfrac{2^2}{5 \cdot 9 \cdot 7 \cdot 5} = \dfrac4{1575} < \dfrac4{400} = \dfrac1{100}$ Since $a_9 < \dfrac1{100}$ and $a_n$ is a decreasing sequence, we have that $a_n < \dfrac1{100}$ for all $n \geq 9$.
One can verify the inequality by hand for $n=9$. After that, incrementing $n$ by $1$ multiplies the result by a factor (significantly) less than $1$. For the numerator is multiplied by $2$, and the denominator is multiplied by at least $11$.