Troubleshooting Alternating Series: Rearranging & Applying L'Hopitals Rule

[kdinser]

Again, my rusty algebra and derivative taking is getting me into trouble.

This is from the section on alternating series. Overall, I think I'm getting the concepts, but some of the solutions to the problems are leaving me scratching my head.

\sum \frac{(-1)^{n+1}(n+1)}{ln(n+1)}

How did the solutions manual go from:

\lim_{n\rightarrow \infty}\frac{(-1)^{n+1}(n+1)}{ln(n+1)}

to this?
\lim_{n\rightarrow \infty}\frac{1}{1/(n+1)}

If someone could just tell me what concept they are using to rearrange this, I'd happily go look it up myself. I dug through my old algebra book and a second calc book and can't find anything like this.

Another problem that I'm having this morning is with an example problem in the same chapter.

They are applying L'Hopital's Rule to test for convergence

\lim_{n\rightarrow \infty}\frac{x}{2^{x-1}}

Again, looking back through past chapters, I can't find a single example of how to take the derivative of
2^{x-1}

I think I'm just forgetting something obvious here.

 

[Doc Al]

Again, looking back through past chapters, I can't find a single example of how to take the derivative of
2^{x-1}

I think I'm just forgetting something obvious here.

\frac{d}{dx} (a^x) = a^x ln(a)

 

[wais]

First of all, don't worry about your rusty algebra and derivative skills. It's completely normal to forget some concepts over time and have to refresh your memory. The important thing is that you are trying to understand and learn.

Now, for the first problem, the solutions manual is using a property of logarithms: ln(a^b) = b*ln(a). Using this property, we can rewrite the denominator ln(n+1) as ln((n+1)^1). Then, we can use the power rule for logarithms to rewrite it as 1*ln(n+1). This allows us to cancel out the (n+1) in the numerator, leaving us with 1/ln(n+1). This is where the 1/(n+1) in the denominator comes from.

For the second problem, you are correct in thinking that we need to take the derivative of 2^{x-1}. This can be done using the chain rule, where we treat 2^{x-1} as the function and x-1 as the input. The derivative of 2^{x-1} is ln(2)*2^{x-1}. Then, we can apply L'Hopital's Rule to get the limit as x approaches infinity of ln(2)*2^{x-1}/1. Since the numerator and denominator both approach infinity, we can use L'Hopital's Rule again to get the limit as x approaches infinity of ln(2)^2*2^{x-1}/1. And since ln(2)^2 is a constant, we can simply ignore it and get the limit as x approaches infinity of 2^{x-1}/1. Finally, we can simplify this to the limit as x approaches infinity of 2^x.

I hope this helps clarify the solutions for you. Keep practicing and don't be afraid to ask for help when needed. Good luck!