Understanding the Radius of Convergence for E 1/n^x in Calculus

incus
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Having a hard time with this one: E 1/n^x , have tried too use n^-x=e^(-x ln n) which in turn e^(...) = lim n->OO (1-(x ln n)/n)^n and then go on with finding the centre, but I feel I'm far far off. How to get it like E an(x-c)^n and use the more straight foreward path.
 
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Hi incus!

The series

\sum_{n=0}^{\infty}\frac{1}{n^x}

is not a power series, so it does not have a radius convergence. It does however have a region of convergence (the x so that the series converges). Is x supposed to be a real or complex variable?

In the case where x is a real variable you can use the integral test to find the region of convergence.

The case where x is complex can be reduced to the real case by considering the real part of x and the absolute value of the terms in the series.
 
You can't ask for a radius of convergence unless you say which point you are expanding around. I'm guessing the actual question is 'for what values of x does the series converge'. Is x complex? Hint: your series is a p-series. And your series defines part of the Riemann zeta function.
 
Thanks for steering me in the right direction yyat and Dick. Got blinded by the question.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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