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!
 A: 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.
A: 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.
A: 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 $m\to+\infty$.)
A: 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 
$$ \sum_{k=m+1}^\infty b^{k^a} < \sum_{k=m+1}^\infty b^{c \ln(k)}
= \sum_{k=m+1}^\infty k^{-c \ln(1/b)} < \int_m^\infty t^{-c \ln(1/b)}\; dt = \dfrac{m^{-c \ln(1/b)+1}}{-c \ln(1/b) + 1}$$  
