How does one prove that

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Can this be extended to higher powers such as:

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closed as unclear what you're asking by user147263, Claude Leibovici, Hans Lundmark, Najib Idrissi, wythagoras Jun 7 '15 at 7:28

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  • $\begingroup$ Could you please clarify what $k$ is in your question? $\endgroup$ – OnceUponACrinoid Jun 7 '15 at 4:12
  • $\begingroup$ k is any real constant, independent of n. $\endgroup$ – sigma Jun 7 '15 at 4:13
  • 1
    $\begingroup$ It's really hard to prove stuff like that. False statements are tricky that way. $\endgroup$ – Joffan Jun 7 '15 at 4:21
  • $\begingroup$ Think about when $k$ is a big positive number. The $n=1$ term is huge on the left and small on the right. Further terms are even more so, so the claimed equality is false. Why do you believe it is true? -1 $\endgroup$ – Ross Millikan Jun 7 '15 at 4:39

The two sides are trivially equal (and diverge) when $k = 1$. Otherwise, they are not equal.

Using the ratio test, we can see that $\sum_{n=0}^{\infty}nk^n$ converges if $|k|<1$ and diverges if $|k| > 1$. Similarly, $\sum_{n=0}^{\infty}nk^{-n}$ converges if $|k| > 1$ and diverges if $|k| < 1$. The two sides clearly cannot be equal if one side converges exactly at the times when the other diverges.

You can make a similar argument for the second equation.

  • $\begingroup$ Minor nitpick: there is equality when $k = 0$. Edit: I suppose its actually not defined for $k = 0$. $\endgroup$ – dalastboss Jun 7 '15 at 4:26

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