Weierstrass M-test

**tbyou87** · December 10th 2007, 09:08 PM

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

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

Thanks

**ThePerfectHacker** · December 11th 2007, 06:14 AM

Originally Posted by tbyou87

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

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

Thanks

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

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