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S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Apr 17, 2007 #2 Re: Power Series Hello, Lisa! \(\displaystyle \L \frac{(-1)(x\,-\,1)(k^2\,-\,1)}{(k\,+\,1)^2\,-\,1}\) Click to expand... The Ratio Test uses absolute value, so we can drop the (−1)\displaystyle (-1)(−1). We have: \(\displaystyle \L\:\left|(x\,-\,1)\cdot\frac{k^2\,-\,1}{k^2\,+\,2k}\right|\) Divide top and bottom by k2 ∣(x − 1)⋅1 − 1k21 + 2k∣\displaystyle k^2\;\;\;\left|(x\,-\,1)\cdot\frac{1\,-\,\frac{1}{k^2}}{1\,+\,\frac{2}{k}}\right|k2∣∣∣∣∣(x−1)⋅1+k21−k21∣∣∣∣∣ Now take the limit as k→∞\displaystyle k\to\inftyk→∞
Re: Power Series Hello, Lisa! \(\displaystyle \L \frac{(-1)(x\,-\,1)(k^2\,-\,1)}{(k\,+\,1)^2\,-\,1}\) Click to expand... The Ratio Test uses absolute value, so we can drop the (−1)\displaystyle (-1)(−1). We have: \(\displaystyle \L\:\left|(x\,-\,1)\cdot\frac{k^2\,-\,1}{k^2\,+\,2k}\right|\) Divide top and bottom by k2 ∣(x − 1)⋅1 − 1k21 + 2k∣\displaystyle k^2\;\;\;\left|(x\,-\,1)\cdot\frac{1\,-\,\frac{1}{k^2}}{1\,+\,\frac{2}{k}}\right|k2∣∣∣∣∣(x−1)⋅1+k21−k21∣∣∣∣∣ Now take the limit as k→∞\displaystyle k\to\inftyk→∞
