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How I find the sum of this series? [math]\displaystyle \sum_{k=1}^{\infty} \frac 1 k \ln(\frac 1 k + 1)[/math]
Okay, that makes sense. What's the next step?
We can then use integration by parts to evaluate the series. Let u = 1/k and dv = ln(1 + 1/k) dk.