I want to know what the steps are to determine whether the following series converge 1) pointwise and 2) uniformly. And how to show if it converges(or doesn't). $\sum \frac{1}{x^k+1}$ on the interval (1,infinity) I've tried on myself for hours but couldn't find a way (tried M-weierstrass but that didn't work out well)

  • $\begingroup$ Pointwise is straightforward enough, it should not be difficult to give a simple upper bound on the sum of absolute values in order to deduce absolute convergence (assuming you intend $k \geq 0$). Work this out and then ping me again about the uniformity part if you're still stuck on it. $\endgroup$ – Ian Mar 26 '17 at 0:58
  • $\begingroup$ 1/(x^k +1) < 1/(x^k)= (1/x)^k is a geometric series so it converges, right? $\endgroup$ – Hasan Aks Mar 26 '17 at 1:07
  • $\begingroup$ That is what I had in mind, yes. $\endgroup$ – Ian Mar 26 '17 at 1:09
  • $\begingroup$ But the step to uniform convergence is the one I totally lose my mind. I tried taking the derivative but I end up getting that the optimal value for x is 0 ( that is not in the domain) and the domain is closed so filling in x=1 to get a M-weierstrass upper bound is also not possible $\endgroup$ – Hasan Aks Mar 26 '17 at 1:13
  • $\begingroup$ So you did an upper bound comparison to get pointwise convergence. But the sum that you compared to does not converge uniformly (indeed, you can use the explicit formula for the partial sums to see this). So that doesn't give uniform convergence. To get a failure of uniform convergence, you would want to give a lower bound on the tail $\sum_{k=N}^\infty \frac{1}{x^k+1}$. Ideally that lower bound would be just a bit smaller than $\sum_{k=N}^\infty x^{-k}$, like say $\sum_{k=N}^\infty x^{-k}/2$. Can you prove a bound like that eventually holds? $\endgroup$ – Ian Mar 26 '17 at 1:24

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