For $q\in (0.5,1)$ why is there $c>0$ such that $f_k(x):=kx^k \leq c q^k$ for all $k\in \mathbb{N}$ In a question I am doing, I am considering the functions $$f_k(x) :=  kx^k$$ ($k\in \mathbb{N}$) on the interval $(-0.5,0)$.
In the solution it says:
pick $q \in (0.5,1)$ and then there exists a constant $c>0$ such that for $|x|\leq 0.5$
$$f_k(x)\leq cq^k$$ for all $k$.
why is $q$ chosen in this specific range and how do we know such a constant $c$ exists?
 A: We have $f'_k(x)=k^2x^{k-1}$.

*

*If $k$ is even, then we see that $f'_k(x)\lt 0$ for $-0.5\le x\lt 0$ and that $f'_k(x)\gt 0$ for $0\lt x\le 0.5$. So, we get $f_k(x)\le k(\pm 0.5)^k=k(0.5)^k$.


*If $k$ is odd, then we see that $f'_k(x)\gt 0$. So, we get $f_k(x)\le k(0.5)^k$.
So, for every $k$ and for $|x|\le 0.5$, we get $$f_k(x)\le k(0.5)^k$$
Finally, we can take $c=k$ to have
$$f_k(x)\le k(0.5)^k\le cq^k$$


why is $q$ chosen in this specific range

I think that the reason why $0.5\lt q$ is that if $0.5\lt q$, then we can easily get $c$ such that $f_k(x)\le k(0.5)^k\le cq^k$.
We can have
$$|S_n|=\bigg|\sum_{k=1}^{n}f_k(x)\bigg|\le \sum_{k=1}^{n}|f_k(x)|\le \sum_{k=1}^{n}kq^k=\frac{q(nq^{n+1}-(n+1)q^n+1)}{(1-q)^2}$$
if $q\not=1$. (If $q=1$, then $|S_n|\le \sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ which depends on $n$.)
I think that the reason why $q\lt 1$ is that if $q\lt 1$, then we can say that
$$|S_n|\le\frac{q(nq^{n+1}-(n+1)q^n+1)}{(1-q)^2}\lt \frac{q}{(1-q)^2}$$
which is independent of $n$ where we have used $nq^{n+1}-(n+1)q^n+1\lt 1$ (which is equivalent to $(1-q)n+1\gt 0$ which is true).
