If $a_n = \sum^n_{r=0} \frac{(\ln10)^n}{r! (n-r)!}$ for $n \geq 0$ .....

Question

Problem:

If $a_n =\sum^n_{r=0} \frac{(\ln10)^n}{r! (n-r)!}$ for $n \geq 0$ then find the value of $a_0+a_1+a_2+\cdots \infty$ My approach:

$a_n = \sum^n_{r=0} \frac{(\ln10)^n}{r! (n-r)!}$

$= \frac{(\ln10)^n}{n!}\sum^n_{r=0} ^nC_r$

$= \frac{(\ln10)^n}{n!} 2^n $ [ Using result $\sum^n_{r=0} ^nC_r = 2^n $]

So, $a_0+a_1+a_2+\cdots \infty = \sum^{\infty}_{n=0} \frac{(2\ln10)^n}{n!} $

Now how to proceed further please suggest. Thanks

Answer 1

Recall that $$1+\frac{x}{1!} +\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=e^x.$$ Using the above series, we should get that $a_0+a_1+a_2+\cdots=100$.