3
$\begingroup$

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?

$\endgroup$
  • 7
    $\begingroup$ $\sum {1\over 2^k}$ converges. If your series converged, then what could you say? $\endgroup$ – David Mitra Aug 27 '12 at 22:52
6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\frac{1}{k}-\frac{1}{2^k}\ge \frac{1}{k}-\frac{1}{2k}=\frac{1}{2k}$.

$\endgroup$
  • $\begingroup$ Nitpickery: doesn't the series have to converge absolutely for that rearrangement to work correctly? $\endgroup$ – Steven Stadnicki Aug 27 '12 at 23:04
  • 1
    $\begingroup$ @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 $\sum_{1}^n b_k$. $\endgroup$ – André Nicolas Aug 27 '12 at 23:12
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.