# [SOLVED] Limit of Truncated Taylor Series to Power.

• May 6th 2010, 12:33 PM
lvleph
[SOLVED] Limit of Truncated Taylor Series to Power.
I am trying to figure out a limit, but am getting a bit stuck.
$
\lim_{n\to \infty} \left(1 + a\frac{t}{n} + a^2\frac{t^2}{2n^2} + a^3\frac{t^3}{3!n^3} + a^4\frac{t^4}{4!n^4}\right)^n
$

Obviously,
$
\lim_{n\to \infty} \left(1 + a\frac{t}{n}\right)^n = e^{at}
$

but I am having trouble extending this.
• May 6th 2010, 01:36 PM
Opalg
Quote:

Originally Posted by lvleph
I am trying to figure out a limit, but am getting a bit stuck.
$
\lim_{n\to \infty} \left(1 + a\frac{t}{n} + a^2\frac{t^2}{2n^2} + a^3\frac{t^3}{3!n^3} + a^4\frac{t^4}{4!n^4}\right)^n
$

Obviously,
$
\lim_{n\to \infty} \left(1 + a\frac{t}{n}\right)^n = e^{at}
$

but I am having trouble extending this.

Squeeze rule: $\bigl(1 + a\tfrac{t}{n}\bigr)^n < \bigl(1 + a\tfrac{t}{n} + a^2\tfrac{t^2}{2!n^2} + a^3\tfrac{t^3}{3!n^3} + a^4\tfrac{t^4}{4!n^4}\bigr)^n < \bigl(e^{at/n}\bigr)^n$. The outer elements both tend to $e^{at}$.
• May 6th 2010, 01:38 PM
lvleph
And we know the RHS holds since the Truncated Taylor Series is an approximation to $e^{at/n}$?
• May 6th 2010, 01:46 PM
Opalg
Quote:

Originally Posted by lvleph
And we know the RHS holds since the Truncated Taylor Series is an approximation to $e^{at/n}$?

I was assuming that the truncated series is less than $e^{at/n}$. But that is only true if at > 0 (which I was taking for granted). If at < 0 then that argument doesn't seem to work.
• May 6th 2010, 01:49 PM
lvleph
Yeah, I was thinking the same thing, but in my problem it is true.
Thank you.