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after the post $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$, I had to ask

Can we telescope the series $$1+\frac{1}{1^k+2^k}+\frac{1}{1^k+2^k+3^k}+\frac{1}{1^k+2^k+3^k+4^k}+\frac{1}{1^k+2^k+3^k+4^k+5^k}+\frac{1}{1^k+2^k+3^k+4^k+5^k+6^k}+\cdots$$ ?

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1 Answer 1

This is not a perfect answer, but a hint. In sum of powers I've a (a bit old) treatize on the Eulerian numbers, which allow, in connection with the binomial-coefficients, to express the sums-of-like-powers as polynomials. Inserting these in the denominators of your sum should allow to do partial sums like in your linked other question. Unfortunately, sum of powers is in german, but the formulae are so simple, that it should be evident, how to proceed on your own. Maybe the same can be found using the wikipedia-article

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What is unfortunate is my level of german not the article being in german. –  Arjang Feb 21 '12 at 10:35
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