elementary number theory - Show $displaystylesum_{k=N_{1}}^{N_{2}-1} k^{s_0}left(frac{1}{k^s}-frac{1}{(k+1)^s}right) rightarrow0$ as $N_1, N_2 rightarrow infty$

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?

Answer

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

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