sequences and series Evaluating $\prod_n=1^\infty\left1+\frac12^n\right$ Mathematics Stack Exchange
The indefinite integral is a rational fraction and is typically solved using partial fractions decomposition.
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- Luckily, the first integral can be evaluated from here but the second integral was the dead-end.
- Both the integrals seem to be hard to evaluate and going into some polylogarithmic forms.
- The indefinite integral is a rational fraction and is typically solved using partial fractions decomposition.
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Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I have an answer to this question(shared Q&A style), but I am curious evaluating business investments as to whether there is another way to evaluate the sum. I have been experimenting with double sums for a few weeks now, so many techniques to attack “simple” sums such as this would be much appreciated! Connect and share knowledge within a single location that is structured and easy to search. I would like to answer the question again with an amusing argument . I don’t want the solution, a pointer towards the right way is sufficient.
- I have an answer to this question(shared Q&A style), but I am curious as to whether there is another way to evaluate the sum.
- While the last link looks promising, I don’t believe that it answers my specific question(but I haven’t read the paper fully, only skimmed).
- I would like to answer the question again with an amusing argument .
- The indefinite integral is a rational fraction and is typically solved using partial fractions decomposition.
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Please try not to evaluate using any expansion series, unless it’s absolutely impossible to solve the limit without using it. While the last link looks promising, I don’t believe that it answers my specific question(but I haven’t read the paper fully, only skimmed). Luckily, the first integral can be evaluated from here but the second integral was the dead-end. Both the integrals seem to be hard to evaluate and going into some polylogarithmic forms.