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How to compute this limit: $\lim_\limits{n\to\infty} \sum_\limits{k=1}^n\frac{1}{1+3^k}$ or written as: $\lim\limits_{n\to\infty}(\frac{1}{1+3^1}+\frac{1}{1+3^2}+\frac{1}{1+3^3}+...+\frac{1}{1+3^n})$

this limit includes computing a summation of series: $\sum_\limits{k=1}^n\frac{1}{1+3^k}$ so that also arouses a more general question: how to compute a summation like $\sum_\limits{k=1}^n\frac{1}{1+a^k}$, where $a=2,3,...,$?

Jose M Serra
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yulingling
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    Welcome to MSE! People are here just to help you,so its better to tell us what have you tried yet? – Math Lover Oct 29 '16 at 06:39
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    There is a "closed form" in terms of $q$-polygamma function. If you want an approximation consider $\sum_{k\geq1}\frac{1}{3^{k}}$. http://mathworld.wolfram.com/q-PolygammaFunction.html – Marco Cantarini Oct 29 '16 at 06:58
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    @VictorBarg It would be better if those same people didn't downvote a beginner. – Jose M Serra Oct 29 '16 at 18:46
  • ISC http://isc.carma.newcastle.edu.au/advancedCalc has the number in its table, but the only evaluation is this series. – GEdgar Oct 29 '16 at 19:10
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    @EnlightenedFunky: Well i think if the OP does not tell us any efforts to solve the problem,then he surely deserves a downvote! – Math Lover Oct 30 '16 at 04:24
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    @VictorBarg Beginners typically don't know any better. – Jose M Serra Oct 30 '16 at 12:32

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