Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.

I got the maclaurin series to be this

$e^x = 1 - 2\frac{x}{1!} + 2^2\frac{x^2}{2!} + \cdots$

$$\sum_{n=0}^{\infty} \frac{(-1)^n (2)^n x^n}{n!}$$

Using the ratio test to find the interval of convergence:

$$\lim_{n\to\infty} \left| \frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} \frac{n!}{(-1)^n 2^n x^n} \right| = 2|x| \lim_{n\to\infty} \frac{1}{n+1} = 2|x|(0) < 1$$

Therefore for any value of x is in the interval of convergence. Is this right?

  • 1
    $\begingroup$ Yes, you're right $\endgroup$ – caverac Apr 19 '17 at 18:57
  • $\begingroup$ Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $\infty$ or zero. If its $\infty$ instead of 0, so like $2|x|\infty > 1$ Can I say there is no point in the interval that is convergence? $\endgroup$ – Tinler Apr 19 '17 at 18:57
  • $\begingroup$ The limit is 0 regardless of the value of $x$, so no need to check anything else $\endgroup$ – caverac Apr 19 '17 at 18:58
  • $\begingroup$ There are no endpoints to check. The interval of convergence is $(-\infty,\infty)$. $\endgroup$ – John Wayland Bales Apr 19 '17 at 18:59
  • $\begingroup$ What happens if its $2|x|\infty > 1$ ?. As in instead of 0 i put infinity. $\endgroup$ – Tinler Apr 19 '17 at 19:05

The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.


This Maclaurin series is valid for $|x|<\infty$.

You know this because the series is valid over that domain. There are no points where the function "blows up".

By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:

  1. Series converges only at $x=0$.
  2. Series converges on an interval centered at $x=0$.
  3. Series converges for all values of $x\in\mathbb{R}$.

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