Showing Continuous Function: Weierstrass Comparison

[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...

 

[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} &lt; 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}