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I have the proof but i don't understand one part. The proof (for $a>1$) goes as follows:

there exists $k \in N$ such that$a<k, \frac{a}{k}<1$ and since $\frac{a}{k+i}<\frac{a}{k}, i \in N$ $$\frac{a^n}{n!}=\frac{a}{1} \cdot \frac{a}{2} \cdot...\cdot\frac{a}{k}\cdot\frac{a}{k+1}\cdot...\cdot\frac{a}{n}<\frac{a^k}{k!}(\frac{a}{k})^{n-k}=\frac{a^k}{k!}\cdot \frac{a^na^{-k}}{k^nk^{-k}}$$

$$0<\frac{a^n}{n!}<\frac{k^k}{k!}(\frac{a}{k})^n$$

then $$0 \leq \lim_{n\to\infty}{\frac{a^n}{n!}}\le\frac{k^k}{k!}\lim_{n\to\infty}{(\frac{a}{k})^n}$$

Now, the part I don't understand is in the expansion of $\frac{a^n}{n!}$. How does one come with $<\frac{a^k}{k!}(\frac{a}{k})^{n-k}$ part, to be more precise, how does one come up with $(\frac{a}{k})^{n-k}$ part of the inequality?

Stefan Stipanovic
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3 Answers3

3

We are given :$$\frac { a^{ n } }{ n! } =\overset { k }{ \overbrace { \frac { a }{ 1 } \cdot \frac { a }{ 2 } \cdot ...\cdot \frac { a }{ k } } } \cdot \overset { n-k }{ \overbrace { \frac { a }{ k+1 } \cdot ...\cdot \frac { a }{ n } } } $$ Now you know:$$\frac { a }{ k+1 } <\frac { a }{ k } \\ \frac { a }{ k+2 } <\frac { a }{ k } $$ $$.$$ $$.$$ $$.$$ $$ \frac { a }{ n } <\frac { a }{ k } $$ so

$$\\ \\ \\ \frac { a^{ n } }{ n! } =\overset { k }{ \overbrace { \frac { a }{ 1 } \cdot \frac { a }{ 2 } \cdot ...\cdot \frac { a }{ k } } } \cdot \overset { n-k }{ \overbrace { \frac { a }{ k+1 } \cdot ...\cdot \frac { a }{ n } } } <\overset { k }{ \overbrace { \frac { a }{ 1 } \cdot \frac { a }{ 2 } \cdot ...\cdot \frac { a }{ k } } } \overset { n-k }{ \overbrace { \frac { a }{ k } \cdot \frac { a }{ k } ...\frac { a }{ k } } } =\frac { { a }^{ k } }{ k! } { \left( \frac { a }{ k } \right) }^{ n-k }$$

haqnatural
  • 22,026
2

We consider $\frac{a}{1}.\frac{a}{2}\ldots \frac{a}{k}\frac{a}{k+1}\ldots \frac{a}{n-2}\frac{a}{n-1}\frac{a}{n}$ Now since $\frac{a}{n}<\frac{a}{n-1}$ and $\frac{a}{n-1}<\frac{a}{n-2}$ and so on till we reach $\frac{a}{n-(n-k)}<\frac{a}{n-(n-k-1)}$ i.e. $\frac{a}{k}<\frac{a}{k-1}$. When we repeatedly use this inequality, we finally will get $\left(\frac{a}{k}\right)^{n-k}$ because there are $n-k$ such terms.

Shahid M Shah
  • 349
1

Here is another proof, maybe simpler: you can use the following basic property of infinite series: if $\sum_{n=1}^{\infty}c_n$ converges, then $\lim_{n\to\infty}c_n=0$. Consider the series $$ \sum_{n=1}^{\infty}\overbrace{\frac{a^n}{n!}}^{c_n} $$ This series converges since by d'Alembert's ratio test $$ L=\lim_{n\to\infty}\frac{c_{n+1}}{c_{n}}=\lim_{n\to\infty}\frac{a}{n+1}=0 $$ and since $L<1$ it follows that $\lim_{n\to\infty}c_n=0$, as requared.

boaz
  • 5,551