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For starters I've noticed that in this case we can apply L'Hôpital's rule. Due to the fact that we get $\left[\frac{\infty}{\infty}\right]$ symbol (or at least I think that we can do this here)

$f(x) =$ $ \sum_{n = 0}^\infty a_{n}\frac{x^n}{n!}$

the first derivative of the numerator equals to:

$f'(x) =$ $a_1+2a_2\frac{x}{2!}+3a_3\frac{x^2}{3!}+...+na_n\frac{x^{n-1}}{n!}$

by repeating this step I end up with something like this:

$f^{n}(x) =a_n$

and the entire limit comes down to:

$\lim_{x \to \infty} \frac{a_n}{e^{x}} = 0$

but from then on I was stuck.

Bartek Żamojtel
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1 Answers1

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By the definition of limit of a sequence, for each $\varepsilon>0$ there is $N$ such that $$ (g-\varepsilon) \sum_{n=N}^\infty \frac{x^n}{n!} \le \sum_{n=N}^\infty \frac{a_n x^n}{n!} \le (g+\varepsilon) \sum_{n=N}^\infty \frac{x^n}{n!} $$ you have

$$ \dfrac{\sum_{n=0}^\infty a_n \frac{x^n}{n!}}{e^x}= \dfrac{\sum_{n=0}^{N-1} a_n\frac{x^n}{n!}- (g\pm\varepsilon)\sum_{n=0}^{N-1} \frac{x^n}{n!}}{e^x} + \dfrac{(g\pm\varepsilon)\sum_{n=0}^{N-1}\frac{x^n}{n!}+\sum_{n=N}^\infty a_n \frac{x^n}{n!} }{e^x} $$ i.e. $$ \dfrac{\sum_{n=0}^{N-1} a_n\frac{x^n}{n!}- (g-\varepsilon)\sum_{n=0}^{N-1} \frac{x^n}{n!}}{e^x} + (g-\varepsilon) \le \dfrac{\sum_{n=0}^\infty a_n \frac{x^n}{n!}}{e^x} \le\dfrac{\sum_{n=0}^{N-1} a_n\frac{x^n}{n!}- (g+\varepsilon)\sum_{n=0}^{N-1} \frac{x^n}{n!}}{e^x} + (g+\varepsilon) $$ Taking the limit for $x \to \infty$ $$ (g-\varepsilon) \le \lim_{x \to \infty}\dfrac{\sum_{n=0}^\infty a_n \frac{x^n}{n!}}{e^x} \le (g+\varepsilon) $$ sending now $\varepsilon \to 0$ gives you the value of the limit

Marco
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