# Simple bound for generalized geometric series

Let $b \in (0,1)$, $m\in \mathbb{N}$ and $a>0$. I want to bound $$\sum_{k=m+1}^\infty b^{k^a} \leq c \; b^{m^a},$$ where $c>0$ is independent from $m$.

Is there a simple way of proving this inequality with $c$ not beeing too large?

Edit: $c$ can only be independent from $m$ if $a>1$. Otherwise it has to grow with $m$ at least linearly, maybe even less...

Thanks!

• One can get a nice value for c when a is at least 1. For smaller a it does not look as good. – The Masked Avenger Oct 28 '14 at 15:19
• Can you give some more details on this nice value?. I only get a satisfying bound for $a\in \mathbb{N}$ applying binomial theorem to $(k+m)^a$. How your value work for non-integer $a$? – L. Omat Oct 28 '14 at 16:44
• If a is at least 1, dividing by b^m^a gives you a convergent series which decays as fast or faster than a geometric series, so c is bounded already by 1/(1-b) if I did not mess up. You can also investigate the ratio b^(m+1^a )/b^m^a if you don't mind c depending on m, but you will still get a smaller value for c as m grows. When a is less than 1, it is not clear to me that the series converges. – The Masked Avenger Oct 28 '14 at 16:53
• @The Masked Avenger: If a<1, you may use Cauchy's condensation test to prove the convergence. – Pietro Majer Oct 28 '14 at 19:40
• An integral convergent test works for this problem, isn't it? It gives two-sided inequalities. – Sergei Oct 30 '14 at 10:15

For $a\ge 1$, the quantity $(m+k+1)^a-m^a$ is increasing w.r.to $m\ge0$ (because $x\mapsto x^a$ is convex). So for $0<b<1$, the term $b^{(m+k+1)^a-m^a}$ is decreasing w.r.to $m\ge0$, and so is $\sum_{k\ge0}b^{(m+k+1)^a-m^a}=b^{-m^a} \sum_{k> m}b^{k^a}$. That is, we have the required bound with $$c=c(b):= \sum_{k>0} b^{k^a} ,$$ which is optimal as it gives an equality for $m=0$ .

For $0<a<1$ the answer is already answered in negative by Christian Remling.

(Note that if $0<a<1$, the above inequality are to be inverted, by concavity of $x\mapsto x^a$, and $b^{-m^a} \sum_{k> m}b^{k^a}$ is increasing; moreover $(m+k+1)^a-m^a\le am^{a-1} (k+1)$ so $$b^{-m^a} \sum_{k> m}b^{k^a}=\sum_{k\ge0}b^{(m+k+1)^a-m^a}\ge \sum_{k\ge0}b^{am^{a-1} (k+1)}=\frac{1}{1-b^{ {a}{m^{a-1}}} }-1,$$ that diverges like $m^{1-a}$ as $m\to+\infty$.)

The sum converges for positive $a$ by the integral test, and since the summand is decreasing, the sum is closely approximated by the integral. The integral can be evaluated easily. Since I am lazy, I used Mathematica:

Assuming[0 < b < 1 && a > 0 && y > 0, Integrate[b^(x^a), {x, y, Infinity}]]


To get $$\frac{y E_{\frac{a-1}{a}}\left(-y^a \log (b)\right)}{a}$$ The exponential integral (which is a Laplace transform) is easily estimated (the better the $c$ the harder you have to work, but a reasonable $c$ is easy.

This cannot work for $a<1$: we then have that $(m+m^{1-a})^a = m^a (1+m^{-a})^a = m^a +O(1)$, so the terms of your series show no decay (except for a multiplicative constant) over an interval of length $\gtrsim m^{1-a}$. Thus $\sum_{k>m} b^{k^a} \gtrsim m^{1-a} b^{m^a}$ and $$b^{-m^a} \sum_{k>m} b^{k^a} \to \infty$$ as $m\to \infty$.

If $a\ge 1$, you can just take $c=b/(1-b)$ (estimate by a geometric series), or you can try to do something smarter, as suggested in the other answers.

• This is at odds with the other answers when a is between 0 and 1. Perhaps you could explain or retract? – The Masked Avenger Oct 28 '14 at 18:57
• To explain: the desired condition cannot hold (the tail being bounded by a previous term times a constant c independent of m) as Christian and Pietro point out; the other answers do not address the desired condition, but do show convergence of the whole sum and some indication of the size of the tail sums, which can get arbitrarily large compared to b^m^a. – The Masked Avenger Oct 28 '14 at 19:38
• @TheMaskedAvenger: I don't think there's a direct contradiction; the $c$'s the other answers produce will depend on $m$ for $a<1$. – Christian Remling Oct 28 '14 at 20:18

Note that $k^a/\ln(k)$ is increasing for $k > \exp(1/a)$. Suppose $m \ge \exp(1/a)$ and $(m+1)^a/\ln(m+1) > 1/\ln(1/b)$. Then with $c = (m+1)^a/\ln(m+1)$ we have $$\sum_{k=m+1}^\infty b^{k^a} < \sum_{k=m+1}^\infty b^{c \ln(k)} = \sum_{k=m+1}^\infty k^{-c \ln(1/b)} < \int_m^\infty t^{-c \ln(1/b)}\; dt = \dfrac{m^{-c \ln(1/b)+1}}{-c \ln(1/b) + 1}$$