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Let $0<s<1$ and $$a_i=\left(i+1\right)^s-(i)^s, \ i \in \mathbb{N}.$$ I'm trying to find a lower bound on $(a_i)_{i\in\mathbb{N}}$ of the form

$$ a_i \geq i^k \ \mbox{for large enough i}.$$

That is, it does not have to hold for small $i$, but only eventually.

Of course, we must have $k<0$. I guess I could prove such bound holds for small enough $k$, however I'd like to have some expression for $k$ (which probably depends on $s$, e.g., $k<-1/s$).

Any thoughts?

Many thanks.

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  • $\begingroup$ You can show that $k = s-1$. $\endgroup$ – Calvin Lin Jun 12 '13 at 22:16
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By the Mean Value Theorem, we have $$\frac{(i+1)^s-i^s}{1}=f'(\xi),$$ where $f(x)=x^s$ and $\xi$ lies between $i$ and $i+1$.

That produces a lower bound of $\dfrac{s}{(i+1)^{1-s}}$, which is not quite of the right shape. But for any $k\gt 1-s$, and large enough $i$ (depending on $k$) this is $\gt \dfrac{1}{i^k}$.

Remark: Prompted by the comment of Calvin Lin, we note that $\dfrac{s}{i^{1-s}}$ is an upper bound. Since the ratio of upper bound to lower bound is close to $1$ for $i$ large, one cannot expect to improve on the $k$.

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  • $\begingroup$ You should add that $k=s-1$ is the best that can be done, which follows almost immediately. $\endgroup$ – Calvin Lin Jun 12 '13 at 22:23

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