Infinite series convergence

[hajfajv]

How am I supposed to approach this type of problem?

 

[topsquark]

How am I supposed to approach this type of problem?

You approached it well. Up until the point where you tried to simplify.

[imath]\dfrac{4}{1 - (e^{2x} - 1)} = \dfrac{4}{e^{2x}} = 4 e^{-2x}[/imath]

You don't need the last step, but it looks a bit nicer to me.

However your canceling method needs to be commented on. You are effectively saying
[imath]\dfrac{4}{1 + 4} = \dfrac{3}{1 + 3} = \dfrac{2}{1 + 2} = \dfrac{1}{1 + 1} = \dfrac{1}{2}[/imath]

But clearly 4/(1 + 4) = 4/5. You can only cancel when you have the same factors in both numerator and denominator. So you can cancel the a in [imath]\dfrac{ab}{ac} = \dfrac{b}{c}[/imath] but you can't cancel the a in [imath]\dfrac{ab + 1}{ac}[/imath].

-Dan

 

[BigBeachBanana]

How am I supposed to approach this type of problem?

You found the sum of the geometric series, but that's not what they're looking for. What's the convergence condition for a geometric series. Apply it.

 

[Steven G]

4/1 is not 3. 4-1 = 3
3/1 is not 2. 3-1 = 2
Why would you put yourself through the pain of taking calculus when you struggle with simple arithmetic?