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tunahan

  • 2 years ago

calculate the radius of convergence and the interval of convergence (with examination of the endpoints) \[\sum_{n=1}^{\infty}\frac{3^{n+1}}{n}.x^{n}\] a try... \[|\frac{a_{n+1}}{a_{n}}|=\] \[ \Large \frac{\frac{3^{(n+1)+1}}{n+1}}{\frac{3^{n+1}}{n}}=\] \[\frac{3^{n+2}}{n+1}.\frac{n}{3^{n+1}}=\frac{3}{1}=3\] is it correct and what i need to do additional?

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  1. cinar
    • 2 years ago
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    \[|R|=3 \quad -3\le R \le 3\]

  2. tunahan
    • 2 years ago
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    that means my solution is nearly correct right?

  3. cinar
    • 2 years ago
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    sorry I made a mistake, first find the Limit, in your case L=3 then Radius is |R|=1/L

  4. tunahan
    • 2 years ago
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    why we get Radius 1/L ?

  5. cinar
    • 2 years ago
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    it is definition..

  6. tunahan
    • 2 years ago
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    hmm ok thank you cinar

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