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Does $\displaystyle\sum^\infty_{n=1} \ln(\frac {2^n+1}{2^n})=\sum^\infty_{n=1} \ln(1+\frac 1 {2^n})$ converge?

Condensation, root, ratio and limit comparison tests don’t help, I don’t know what to do… Any hints please?

**Note:** no integral test.

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I disagree that the limit comparison doesn’t help. Since the summands of the series are positive,

$$\lim_{n\to \infty} 2^n \ln\left(1 + \frac{1}{2^n}\right) = \lim_{x\to 0} \frac{\log(1 + x)}{x} = \frac{d}{dx}\bigg|_{x = 0} \log(1 + x) = 1,$$

and $\sum_{n = 1}^\infty \frac{1}{2^n}$ converges, by limit comparison the series $\sum_{n = 1}^\infty \ln\left(1 + \frac{1}{2^n}\right)$ converges.

Since $0< \ln (1+x) < x$ for $x>0$, then

$$ 0 < \sum_{n=1}^\infty \ln (1+2^{-n}) \le \sum_{n=1}^\infty 2^{-n} = 1 $$

Note that $$\ln(1+x) = \sum_{n=1}^\infty (-1)^n \frac{x^n}{n}$$

This means $$\ln(1+x) = x + O(x^2)$$ as $x \to 0$.

Thus $$\sum_{n=1}^\infty \ln(1+2^{-n}) = \sum_{n=1}^\infty \left( \frac{1}{2^n} + O\left(\frac{1}{2^{2n}}\right)\right)$$

Thus $$\sum_{n=1}^\infty \ln(1+2^{-n}) \le \sum_{n=1}^\infty \frac{1}{2^n} + C \sum_{n=1}^\infty \frac{1}{2^{2n}}$$

Both of the sums are convergent on the right. So the original sum converges by the direct comparison test.

Another method is to show that $\lim_{n\to \infty} \frac{\ln(1+2^{-n})}{2^{-n}} = 1$. This follows from the definition of the derivative (or if you like, l’Hopital’s rule).

Thus by the limit comparison test $\sum \ln(1+ 2^{-n})$ and $\sum 2^{-n}$ either both converge or they both diverge. We know the second converges since it is a geometric series, thus the original converges as well.

Recall that, as $x \to 0$,$$

\ln(1+x) \sim x

$$ giving, as $n \to \infty$, $$

\ln\left(1+\frac 1 {2^ n}\right)\sim\frac 1 {2^ n}

$$ and by comparison your series is **convergent**.

Convergence is straightforward by comparison with $\sum_{n\geq 1}\frac{1}{2^n}=1$. Actually

$$ S=\sum_{n\geq 1}\log\left(1+\frac{1}{2^n}\right) = \sum_{m\geq 1}\frac{(-1)^{m+1}}{m}\sum_{n\geq 1}\frac{1}{2^{nm}}=\sum_{m\geq 1}\frac{(-1)^{m+1}}{m(2^m-1)} $$

and

$$ S=\log\left(1+\sum_{n\geq 1}\frac{q(n)}{2^n}\right) $$

where $q(n)$ is the number of restricted partitions of the positive integer $n$ into a sum of distinct positive numbers that add up to $n$ when order does not matter and repetitions are not allowed.

The series is positive and its general term is such that $\log{(1+\frac{1}{2^n})}\sim \frac{1}{2^n}$ and the RHS is the general term of a convergent series. Therefore the series we are looking at is convergent.

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