# Continuous function

akoska
How do I show that f(x)=sum (from n=0 to infinity) cos(nx)e^-nx is a continuous function? x is from (0, infinity)

So, I need to show that the series converges uniformly. I'm trying to say that |cos nx e^-nx| <= |e^-nx| and use Weierstrass comparison, but I can't find a function M_n to use for Weierstrass.

I feel like I'm missing something really simple...

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maverick280857
For positive $n$ what is $e^{-nx}$ always less than?

akoska
For positive $n$ what is $e^{-nx}$ always less than?

1, but the sum of 1s is obviously not convergent.

maverick280857
1, but the sum of 1s is obviously not convergent.

Okay, let's see:

$$f(x) = \sum_{n = 0}^{\infty} e^{-nx}\cos(nx)$$

Clearly,

$$|e^{-nx}\cos(nx)| \leq e^{-nx}$$

Now, as you said, $e^{-nx} < 1$ for all $x \in (0,\infty)$ and nonnegative n. So, clearly, your "$M_{n}$" is simply the (convergent) geometric sum

$$\sum_{n = 0}^{\infty} e^{-nx}$$