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Is there a handy way to tell if $\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{2^k}\right)$ diverges or not? I have a hunch that it diverges, since it looks like the sum is just $\zeta(1)-1=\infty$. But I'm not sure one can rearrange the series as $$ \sum_{k=1}^\infty\frac{1}{k}-\sum_{k=1}^\infty\frac{1}{2^k}.$$

Is that a valid move?

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$\sum {1\over 2^k}$ converges. If your series converged, then what could you say? – David Mitra Aug 27 '12 at 22:52
up vote 6 down vote accepted

Suppose the series $\sum\limits_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{2^k}\right)$ converged. The series $\sum\limits_{k=1}^\infty\frac{1}{2^k}$ converges since it is a geometric series with common ratio $\frac12$. With two convergent series, you can add them term by term to get that $$ \sum_{k=1}^\infty\frac1k\tag{1} $$ converges.

The series in $(1)$ diverges, so the series $\sum\limits_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{2^k}\right)$ diverges.

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It is a valid move if at least one of the series converges. In your case one of them does.

Or else one could note that for $k \ge 2$, $\frac{1}{2^k}\le \frac{1}{2k}$. Thus if $k\ge 2$, then $\frac{1}{k}-\frac{1}{2^k}\ge \frac{1}{k}-\frac{1}{2k}=\frac{1}{2k}$.

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Nitpickery: doesn't the series have to converge absolutely for that rearrangement to work correctly? – Steven Stadnicki Aug 27 '12 at 23:04
@StevenStadnicki: It would certainly not be nitpickery! But one can show that if $\sum b_k$ converges, then $\sum (a_k+b_k)$ converges iff $\sum a_k$ does. This is because there is very good control over the partial sums $\sum_{1}^n b_k$. – André Nicolas Aug 27 '12 at 23:12

Hint: Try a limit comparison test with the divergent harmonic series.

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