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Prove that if $s >1$, then $\displaystyle {\sum_{r=1}^n} \left( {\frac{s}{r}- \frac{1}{r^3}} \right) - s \log n$ tends to a limit as $n \to \infty$ ($s$ being fixed), and that if this limit is $\phi (s)$ then [also show that] $0 \le \left[ {\phi (s) + \frac {1}{s-1}} \right] \le (s-1)$

From A Course of Mathematical Analysis by Shanti Narayan pp.332.

Edit: Solved! $\displaystyle {\lim_{n \to \infty} \sum_{r=1}^n} \left( {\frac{s}{r}- \frac{1}{r^3}} \right) - s \log n$ $= \displaystyle {\lim_{n \to \infty} \sum_{r=1}^n} \left( {\frac{s}{r}} \right) - s \log n -\displaystyle {\lim_{n \to \infty} \sum_{r=1}^n} \left( {\frac{1}{r^3}} \right)$ $=s \gamma - \zeta(3)$ [$\gamma$ := Euler-Mascheroni constant]. Now the inequality, which I suppose to be wrong (but actually it is not), was proved using Mathematical Induction. Thanks to all.

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    @srivatsan Can you please be more specific? From where you got $\zeta(3)$? And what is this $\gamma $ here? Is it $\infty$?2011-09-11

2 Answers 2

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Euler Constant

Here is a nice way to see $\displaystyle\sum_{k=1}^{n-1} \frac{1}{k} - \log n $ converges (which essentially answers the first question). I have shaded in the "error" when estimating $\log n$ with the sum, and then shifted those regions horizontally. Clearly, as we take higher partial sums, the sum of the errors will increase. We can always slide them to fit into the $1$ by $1$ rectangle on the left, so it is bounded as well and hence converges. Now it is your exercise to convert this into a rigorous proof.

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    Yay! Proof by I$n$duction worked. Pro$b$lem solved.2011-09-11
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Hint: Check the definition of the Euler-Mascheroni constant for two terms. The third term converges easily.

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    @guarav: I see. Deleted.2011-09-11