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fix math error
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Anthony Quas
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It converges for $q<1$ and diverges forall $q\ge 1$$q$. You can separate the interval into the ranges $[e,1/\sqrt c]$ and $[1/\sqrt c,\infty)$. On the first range, the exponential term is essentially 1. On the second range, the logarithm is essentially constant.

It converges for $q<1$ and diverges for $q\ge 1$. You can separate the interval into the ranges $[e,1/\sqrt c]$ and $[1/\sqrt c,\infty)$. On the first range, the exponential term is essentially 1. On the second range, the logarithm is essentially constant.

It converges for all $q$. You can separate the interval into the ranges $[e,1/\sqrt c]$ and $[1/\sqrt c,\infty)$. On the first range, the exponential term is essentially 1. On the second range, the logarithm is essentially constant.

Source Link
Anthony Quas
  • 23.2k
  • 5
  • 63
  • 98

It converges for $q<1$ and diverges for $q\ge 1$. You can separate the interval into the ranges $[e,1/\sqrt c]$ and $[1/\sqrt c,\infty)$. On the first range, the exponential term is essentially 1. On the second range, the logarithm is essentially constant.