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This might be ridiculously easy but I just forgot about series.

Consider the series $\sum_{k=1}^\infty \frac{1}{k^2-2}$. Does it converge? What about $\sum_{k=1}^\infty \frac{1}{k^2-r}$ for any $r>0$?

I tried ratio test and it's inconclusive. Direct comparison... I compared it to a p-series and didn't get anything.

Thanks

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  • $\begingroup$ How can direct comparison don't work? Note that $\frac{1}{k^2+r}\leq\frac{1}{k^2},\forall r>0$ and $\sum_{k=1}^\infty\frac{1}{k^2}<\infty$. $\endgroup$ – DiegoMath Oct 31 '14 at 2:51
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Well, for all partial sums, $$\sum_{k=1}^n \frac{1}{k^2+2} < \sum_{k=1}^n \frac{1}{k^2} $$ and the second series does converge. Also, the original series is monotonically increasing. Hence ...

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  • $\begingroup$ I apologize it was a minus there... $\endgroup$ – verticese Oct 31 '14 at 2:53
  • $\begingroup$ Even so, just shift the index after the first couple of terms and the same type of bound holds: $1/(k^2 - 2) < 1/(k-2)^2$ for all $k \geq 3$. $\endgroup$ – Simon S Oct 31 '14 at 2:56
  • $\begingroup$ So in the same spirit $\sum_k^\infty \frac{1}{k-r}$ is convergent because you can shift backwards by $r$? $\endgroup$ – verticese Oct 31 '14 at 2:59
  • $\begingroup$ $1/(k^2 - r)$, yes, just work through the algebra to make sure you've got the appropriate lower index. Then just parcel out those finite number of terms before that index separately. $\endgroup$ – Simon S Oct 31 '14 at 3:03

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