Gold Member

MHB

POTW Director

Last edited: Aug 22, 2022

If (a_n\to a), write
[
\sum_{n\ge0}a_n\frac{x^n}{n!}
=ae^x+\sum_{n\ge0}(a_n-a)\frac{x^n}{n!}.
]
Since (a_n-a\t, the second sum is (o(e^x)). Hence
[
\boxed{\sum_{n\ge0}a_n\frac{x^n}{n!}\sim ae^x\quad(x\to\infty).}
]

mxwlljack said:

If (a_n\to a), write
[
\sum_{n\ge0}a_n\frac{x^n}{n!}
=ae^x+\sum_{n\ge0}(a_n-a)\frac{x^n}{n!}.
]
Since (a_n-a\t, the second sum is (o(e^x)). Hence
[
\boxed{\sum_{n\ge0}a_n\frac{x^n}{n!}\sim ae^x\quad(x\to\infty).}
]

It seems to me that you did not show that ## \sum_{n\ge0}(a_n-a)\frac{x^n}{n!}\sim0 ##, which is crucial to the proof.