Convergence or Divergence of $ 1 – 1 + \frac{1}{2} – . . .$ and another series.

convergence-divergencereal-analysissequences-and-series

I'm testing ideas about the convergence and divergence of series:

Say I have a series $$ 1 – 1 + \frac{1}{2} – \frac{1}{2} + \frac{1}{3} – \frac{1}{3} + . . .$$

I believe that this series would converge to $0$ as every odd term in the series is canceled by the following term. (In addition, the individual terms are decreasing to 0 from 1 and -1.) In other words, the sequence of partial sums of this series tend to a limit, $0$.

Say I have a different series $$1 – \frac{1}{2} + \frac{1}{2} – \frac{1}{2^2} + \frac{1}{3} – \frac{1}{2^3} + \frac{1}{4} – \frac{1}{2^4} + . . .$$

My assumption would be that the sum of odd terms is not canceled out by the sum of even terms. In other words, this series would increase to infinity. The partial sum of the first $i$ terms would be less than the partial sum of the first $i+2$ terms, and so forth.

How do these series converge or diverge? Am I looking at this the right way?

Best Answer

$$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{2^3} + \frac{1}{4} - \frac{1}{2^4} + . . .$$

can be coded as $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big)$. If this series would converge, as the sum of convergent series is a convergent series, then $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big) +\sum_{n\geq1}\frac{1}{2^n}=\sum_{n\geq1}\frac{1}{n}$ would converge. This is a absurd, therefore the series diverges.

The reasoning behind your conclusion - "The partial sum of the first $i$ terms would be less than the partial sum of the first $i+2$ terms, and so forth" might be wrong... take for example $e=\sum_{n\geq0}\frac{1}{n!}$.

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