Consider the following infinite series: ∑k=0∞2k1+1x2k

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Consider the following infinite series:  ∑k=0∞2k1+1x2k

Hattie Schaeffer · Accepted Answer

To me, the sum can be derived informally as  ∑k=0∞2kx2k1+x2k=x∑k=0∞ddx[log⁡(1+x2k)]  =xddx[log⁡(∏k=0∞1+x2k)]  =xddx[log⁡(∏k=0∞1−x2k+11−x2k)]  =−xddx[log⁡(1−x)]  =x1−x  The key is the summand in the original series is proportional to the derivative of logarithm of something simple. Once we turn the sum of log to a log of product and realize the product is a telescoping one, the rest is just following your nose.

usumbiix · Accepted Answer

Let |x|&amp;lt;1 and k≥0. Using the factorization 1−x2k+1=(1−x2k)(1+x2k), we find  2kx2k1+x2k=2kx2k[1+x2k−2x2k](1+x2k)(1−x2k)  =2kx2k1−x2k−2k+1x2k+11−x2k+1  So since limk→∞2kx2k1−x2k=0, the series telescopes to x1−x

karton · Accepted Answer

Let f(x) be the series. Rewriting, we get f(x)=∑k=0∞2kx2k1+x2k Let sn be the n-th partial sum. We then have x1−x−sn=x1−x−x1+x−...−2nx2n1+x2n =2n+1x22n+11+x2n+1 One can show that the above tends to 0 as n→∞ for |x|&amp;lt;1

Consider the following infinite series: \sum_{k=0}^\infty\frac{2^k}{1+\frac{1}{x^{2k}}}

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