Radius of convergence of power series $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$











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Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.




My try:



Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get



$displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or



the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??






sequences-and-series proof-verification power-series alternative-proof







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    up vote
    2
    down vote

    favorite
    1













    Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.




    My try:



    Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get



    $displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or



    the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??






    sequences-and-series proof-verification power-series alternative-proof







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      up vote
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      up vote
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      Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.




      My try:



      Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get



      $displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or



      the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??






      sequences-and-series proof-verification power-series alternative-proof







      share|cite|improve this question














      Find radius of convergence of $displaystyle sum_{n=1}^{infty}frac{x^{6n+2}}{(1+frac{1}{n})^{n^2}}$.




      My try:



      Using Cauchy's Root test for convergence $displaystyle lim_nsup sqrt[n]{|a_n|}|(z-c)|<1$, we get



      $displaystyle lim_nsup sqrt[n]{|frac{1}{Big(1+frac{1}{n}Big)^{n^2}}|},|x^6|<1implies lim_nsup frac{1}{Big(1+frac{1}{n}Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or



      the interval of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$ and radius of convergence is $2e^frac{1}{6}$. Am I correct ??






      sequences-and-series proof-verification power-series alternative-proof




      sequences-and-series proof-verification power-series alternative-proof






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      asked yesterday









      Yadati Kiran

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          That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.






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          • ok yeah ! I calculated the length of the interval. Thanks @Fred.
            – Yadati Kiran
            yesterday











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          active

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          active

          oldest

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          up vote
          1
          down vote



          accepted










          That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.






          share|cite|improve this answer





















          • ok yeah ! I calculated the length of the interval. Thanks @Fred.
            – Yadati Kiran
            yesterday















          up vote
          1
          down vote



          accepted










          That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.






          share|cite|improve this answer





















          • ok yeah ! I calculated the length of the interval. Thanks @Fred.
            – Yadati Kiran
            yesterday













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.






          share|cite|improve this answer












          That the intervall of convergence is $displaystyle Big(-e^frac{1}{6},e^frac{1}{6}Big)$, is correct. But the radius of convergence is $e^frac{1}{6}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Fred

          41.5k1641




          41.5k1641












          • ok yeah ! I calculated the length of the interval. Thanks @Fred.
            – Yadati Kiran
            yesterday


















          • ok yeah ! I calculated the length of the interval. Thanks @Fred.
            – Yadati Kiran
            yesterday
















          ok yeah ! I calculated the length of the interval. Thanks @Fred.
          – Yadati Kiran
          yesterday




          ok yeah ! I calculated the length of the interval. Thanks @Fred.
          – Yadati Kiran
          yesterday


















           

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