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Show $\displaystyle\sum_{k=N_{1}}^{N_{2}-1} k^{s_0}\left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right) \rightarrow0$ as $N_1, N_2 \rightarrow \infty$, where $s \gt s_0 \ge0$ and $k, N_1, N_2 \in \mathbb{N}.$

I'm going through some number theory notes a friend gave me to learn more about this field of mathematics, and they show the following proof (on the way to proving something in relation to the abscissa of convergence), which I think is incorrect. From the statement above, they argue that since $k, s_0 \ge 0$, it follows that $k^{s_0} \leq (k+1)^{s_0}$, so that (and here, I think, we have a non-sequitur):

$\displaystyle\sum_{k=N_{1}}^{N_{2}-1} k^{s_0}\left(\frac{1}{k^s}-\frac{1}{(k+1)^s}\right) \leq \displaystyle\sum_{k=N_{1}}^{N_{2}-1} \left(\frac{1}{k^{s-s_{0}}}-\frac{1}{(k+1)^{s-s_{0}}}\right) = \frac{1}{{N_1}^{s-s_0}} - \frac{1}{{N_2}^{s-s_0}} \rightarrow0$ as $N_1, N_2 \rightarrow \infty$.

Am I wrong in thinking the first inequality doesn't hold, as we're subtracting a greater quantity on the RHS rather than a smaller one?

If so, how would you prove the original statement?

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The inequality is applied termwise. You could directly check if the inequality holds termwise. Is it true that $$ k^{s_0} \left( \frac 1{k^s} - \frac 1{(k+1)^s} \right) \le \frac 1{k^{s-s_0}} - \frac 1{(k+1)^{s-s_0}} $$ ? You should compute that yourself. I think one good question you need to ask yourself is "what does it mean for $N_1, N_2$ to go to infinity. When you have only one variable it makes sense ; when it's large enough, you can be as precise as you want, and that's what a limit means. When you have two variables, you can make them go to infinity in many different ways. –  Patrick Da Silva Aug 2 '13 at 20:43
    
@PatrickDaSilva, just plugging in $k = 2, s_0 = 1$ and $s = 2$ shows the inequality to be untrue, as I suspected. As for your question, I thought of that myself. I don't like this is how the notes go about it, but I'm just trying to work within what's given there. And I guess if the argument provided were true, then there would be no problem with having two variables go to infinity, as they would appear separately. However, I'm having trouble showing the original statement, since without that false inequality I can't decouple $N_1$ and $N_2$, and stumble upon the exact problem you allude to. –  Ryker Aug 2 '13 at 20:51
    
I didn't mean plugging in examples, that is not 'thinking'. I meant showing that the inequality is equivalent to $$ - \frac{k^{s_0}}{(k+1)^s} \le -\frac{(k+1)^{s_0}}{(k+1)^s} $$ which after cancelling the $(k+1)^s$'s is clearly always false, in the sense that your inequality in the proof you've quoted should be reversed. –  Patrick Da Silva Aug 2 '13 at 20:59
    
@PatrickDaSilva, I know, this is how I did it when I first suspected there was something wrong with the proof. But since it was in the notes, I figured I'd ask for help on how to actually show the original statement is true without this false inequality. –  Ryker Aug 2 '13 at 21:02
    
I understand, I was commenting but I'm thinking too. –  Patrick Da Silva Aug 2 '13 at 21:04
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1 Answer 1

up vote 1 down vote accepted

Yes, you're right, since $k^{s_0} < (k+1)^{s_0}$ for $s_0 > 0$ [the inequality is correct for $s_0 = 0$], you then actually have

$$k^{s_0}\left(\frac{1}{k^s} - \frac{1}{(k+1)^s}\right) > \frac{1}{k^{s-s_0}} - \frac{1}{(k+1)^{s-s_0}}.$$

The statement can be proved quite easily (I hesitate to say most easily) by writing $\frac{1}{k^s} - \frac{1}{(k+1)^s}$ as an integral:

$$k^{s_0}\left(\frac{1}{k^s} - \frac{1}{(k+1)^s}\right) = k^{s_0} \int_k^{k+1} \frac{s}{t^{s+1}}\, dt \leqslant \int_k^{k+1} \frac{s}{t^{s-s_0+1}}\,dt.$$

Now we can estimate

$$\sum_{k = N_1}^{N_2-1} k^{s_0}\left(\frac{1}{k^s} - \frac{1}{(k+1)^s}\right) \leqslant \int_{N_1}^{N_2} \frac{s}{t^{s-s_0+1}}\,dt = \frac{s}{s-s_0}\left(\frac{1}{N_1^{s-s_0}} - \frac{1}{N_2^{s-s_0}}\right).$$

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Thanks, that was brilliant. –  Ryker Aug 3 '13 at 0:10
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