Proving the Limit of an Infinite Sum

[sarrah1]

Prove that$\lim_{{n}\to{\infty}}\sum_{j=0}^{n} {n \choose j} \frac{{(x-a)}^{n+j}}{(n+j) !} = 0 $
thanks
Sarrah

 

[Euge]

Hello again Sarrah,

To solve this problem, it suffices to show that

$$\lim_{n\to \infty} \sum_{j = 0}^n \binom{n}{j}\frac{|x - a|^{n+j}}{(n+j)!} = 0.$$

To this end, let $f_n(x;a)$ be the summation expression. Then

$$0 \le f_n(x;a) \le \frac{|x - a|^n}{n!}\sum_{j = 0}^n \binom{n}{j}|x - a|^j = \frac{M^n}{n!},$$

where $M = |x - a|(1 + |x - a|)$. Since $M \ge 0$, $\lim\limits_{n\to \infty} \frac{M^n}{n!} = 0$. Hence, by the squeeze theorem, $\lim\limits_{n\to\infty} f_n(x;a) = 0$.

 

[sarrah1]

Dear Euge

Always there when i am in need of help
very grateful
sarrah