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!



Simple bound for generalized geometric series
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 $m\to+\infty$.)


