- #1
jhudson1
- 16
- 0
This problem is taken from the visual calculus archives:
http://archives.math.utk.edu/visual.calculus/6/series.15/index.html
Determine whether the series is convergent:
infinity
Ʃ (1-2^n)/(1+2^n)
n=0
Checking this on WolframAlpha I see that the limit of the series is -1, so the series is divergent by the Nth term test. However, I don't understand why the limit is -1.
The explanation given by WolframAlpha seems to rely on some algebraic manipulation which does not immediately jump out at me. Is the process by which WolframAlpha arrived at the answer the most efficient? Should I have seen that? Or is there a better way?
http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+(1-2^n)/(1+2^n)
http://archives.math.utk.edu/visual.calculus/6/series.15/index.html
Determine whether the series is convergent:
infinity
Ʃ (1-2^n)/(1+2^n)
n=0
Checking this on WolframAlpha I see that the limit of the series is -1, so the series is divergent by the Nth term test. However, I don't understand why the limit is -1.
The explanation given by WolframAlpha seems to rely on some algebraic manipulation which does not immediately jump out at me. Is the process by which WolframAlpha arrived at the answer the most efficient? Should I have seen that? Or is there a better way?
http://www.wolframalpha.com/input/?i=lim+n+approaches+infinity+(1-2^n)/(1+2^n)