So thanks to tph we know that $\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$ converges on $\displaystyle (-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$.

To use the Weierstrass M-test, how do we find a bound for that sum.

Thanks

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- Dec 10th 2007, 09:08 PMtbyou87Weierstrass M-test
So thanks to tph we know that $\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$ converges on $\displaystyle (-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$.

To use the Weierstrass M-test, how do we find a bound for that sum.

Thanks - Dec 11th 2007, 06:14 AMThePerfectHacker
What are you trying to show? We can show that if $\displaystyle 0<r<R$ where $\displaystyle R=\sqrt{3/2}$ then $\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$ converges uniformly on $\displaystyle [-r,r]$ by choosing $\displaystyle M_k = \frac{k2^k}{3^k}r^{2k}$.