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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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The $r^{-3}$ is in some sense a red herring. Try solving the problem ignoring the term entirely (hint: think integrals/Riemann sums). Then put the term back in and see what changes are necessary. –  Aaron Sep 11 '11 at 4:27
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I calculated the sum of the series to be $\phi(s) := s \gamma + \zeta(3)$. I could prove the left hand side inequality, but the right hand side inequality seems wrong. Specifically, as $s \to 1$, $\phi(s)$ is bounded from below, so that $\phi(s) + \frac{1}{s-1}$ is diverges to $\infty$ as $s \to 1$. In contrast, $s-1$ converges to $0$. So the inequality cannot hold as $s \to 1$. –  Srivatsan Sep 11 '11 at 5:15
    
@Srivatsan Did you notice that $s>1$ ? –  gaurav Sep 11 '11 at 5:25
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@gaurav Yes, I did. Does that change anything? [Oh, btw $\phi(s) = s\gamma - \zeta(3)$; the formula I mentioned has a typo. That does not change my conclusion however.] –  Srivatsan Sep 11 '11 at 5:26
    
@srivatsan Can you please be more specific? From where you got $\zeta(3)$? And what is this $\gamma $ here? Is it $\infty$? –  gaurav Sep 11 '11 at 5:55
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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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Thanks. @Srivatsan, in chat, gave full details about how to solve it. But I'm having problem proving the 'inequality'. –  gaurav Sep 11 '11 at 12:40
    
Yay! Proof by Induction worked. Problem solved. –  gaurav Sep 11 '11 at 12:49
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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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Function of $s$ & the limit of summation. –  gaurav Sep 11 '11 at 4:25
    
@guarav: I see. Deleted. –  Ross Millikan Sep 11 '11 at 4:30
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