Is the series sum from n=1 to infinity of (-1)^n convergent or divergent? Justify your answer. Can you now resolve the difficulty of the following:Divergent series:

S=1+1/2+1/4+1/8+1/16+…You have probably seen the following trick to sum this series: if we call the above sum S, then if we multiply by 2, we obtain: 2S=2+1+1/2+1/4+…=2+SHence S=2, so the series sumes to 2. However, if you apply the same trick to the series S=1+2+4+8+16+… one gets nonsensical results:

2s=2+4+8+16+…=S-1 => S=-1So the same reasoning that shows that 1+1/2+1/4+…=2 also gives that 1+2+4+8+…=-1. Why is it that we trust the first equation but not the second? A similar example arises with the series:

S=1-1+1-1+1-1+…;We can write S=1-(1-1+1-1+…)=1-S and that S=1/2

Instead, we can write S= (1-1)+(1-1)+(1-1)+…=0+0+… and hence S=0

Or, we can write S=1+(-1+1)+(-1+1)+…=1+0+0+…and S=1. Which one is correct?I used LaTeX to generate a PDF file for the solution.

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