How does one compute the limit of the sequence:
$$\sum_{k = 0}^{n}\frac{1}{3k+1} - \frac{\ln(n)}{3}$$
I would apreciate a hint.
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It should be trivial to derive the sum from the result in the question that @Lucian referred to above. Accordingly, this question is about to be closed as a duplicate … – Harald Hanche-Olsen Jan 28 '14 at 15:33
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Yes, thank you :D – user124462 Jan 28 '14 at 15:47
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Hint: use a comparison series/integral, by writing $$\frac{\ln n}{3} = \int_{1}^n \frac{dx}{3x}=\sum_{k=1}^{n-1} \int_{k}^{k+1}\frac{dx}{3x}$$ and $$\sum_{k=0}^n \frac{1}{3k+1} = \sum_{k=0}^n \int_{k}^{k+1}\frac{dx}{3k+1}$$
Edit: is not a straightforward approach -- I was thinking about proving convergence.
Clement C.
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1This way it is easy to show that the sequence converges... I don't see how it helps at computing the limit(but of course, I might be missing something - If I am, could I get the complete calculation?) – user124462 Jan 28 '14 at 15:22
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Yes, sorry -- not direct from there. (i was thinking about proving convergence, indeed) – Clement C. Jan 28 '14 at 15:37