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Evaluate $\sum_{n=1}^{\infty }\ln \left (\frac{7^n+1}{7^n} \right )$ .

Found this question on Art of Problem Solving. It was stuck in the "solved" section, but I couldn't find a solution, and I myself am stumped.

Apparently it could also be simplified to $\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k+1}}{k\left ( 7^{k}-1 \right )}$ , but I don't follow this either.

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2  
AOPS say $$\sum_{n=1}^{\infty} \ln \left( \frac{7^n+1}{7^n} \right)$$ –  user17762 Mar 6 '12 at 0:01
3  
@SivaramAmbikasaran: I see. Very strange of OP to miss that! –  Aryabhata Mar 6 '12 at 0:10
    
You'll want $k$ to go from $1$ to $\infty$, not from $0$. –  Robert Israel Mar 6 '12 at 2:51
    
@SivaramAmbikasaran I'm new to LaTeX. Give me a bit of leniency. :P –  badreferences Mar 6 '12 at 17:54

2 Answers 2

This is $$\log \prod_{n=1}^\infty (1 + 7^{-n}) = \log \phi(1/49) - \log \phi(1/7)$$
where $$ \phi(q) = \prod_{n=1}^\infty (1 - q^n) $$ is the Euler function. I doubt that you can get a much simpler "closed form" than that.

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Using the inequality

$$\log (1+x) \ge x - \frac{x^2}{2}$$

we see that the series diverges: $$\log\left(\frac{7n + 1}{7n}\right) \ge \frac{1}{7n} - \frac{1}{98n^2}$$

EDIT:

If the series is $\sum_{n=1}^{\infty}\log\left(\frac{7^n + 1}{7^n}\right)$ (as pointed out in the comments), then, using the Taylor series expansion of $\log(1+x)$:

$$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots $$

and the geometric series sum

$$\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}$$

we get the sum which you state.

(Of course, that would need some justification, but I believe it is doable).

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