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## Putnam 2016 B6

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### Putnam 2016 B6

Fri Dec 09, 2016 10:00 am

Posts: 46
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The following series can be found at this year's Putnam Exam.

Evaluate the series:

$$\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\sum_{n=0}^{\infty}\frac{1}{m2^n+1}$$

### Re: Putnam 2016 B6

Sun Dec 11, 2016 5:47 pm

Posts: 1
Observing that $(1+m2^{n})^{-1}=\int_{0}^{1}x^{m2^{n}}dx$, we can write the wanted series as
$$\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\sum_{n=0}^{\infty}\int_{0}^{1}x^{2^{n}m}dx.$$
Exchanging the double sum with the integral (which is justified because of Fubini's theorem, given the absolute integrability of the series), we get
$$\int_{0}^{1}\sum_{n=0}^{\infty}\left(\sum_{m=1}^{\infty}\frac{(-1)^{m-1}\left(x^{2^{n}}\right)^{m}}{m} \right)dx.$$
The inner sum is just $\log(1+x^{2^{n}})$, so we get
$$\int_{0}^{1}\sum_{n=0}^{\infty}\log(1+x^{2^n})dx = \int_{0}^{1}\log\left(\prod_{n=0}^{\infty}(1+x^{2^{n}}) \right)dx = \int_{0}^{1}\log\left(1+x+x^{2}+x^{3}+\ldots \right)dx = \\ = -\int_{0}^{1}\log(1-x)dx = \left[(1-x)\log(1-x)+x \right]_{0}^{1}=1.$$