Some estimation a series by integral

Question

Let $a \in (0,1)$. Does there exist a constant $C>0$ or function $C(a)>0$, which may be a function of $a$ but not $k$, such that $$ \sum_{n=1}^\infty a^n n^k \leq C(a) \int_0^\infty a^x x^k dx $$ for all $k \in \mathbb N$?

Answer 1

Recalling the Mellin transform

$$ F(s)=\int_{0}^{\infty} x^{s-1} f(x) dx. $$

Our integral is nothing but the Mellin transform of the function $a^x=e^{\ln(a)x}$. Now, if $a \in (0,1) $ then $\ln(a) < 0 $ and we have

$$ \int_{0}^{\infty} a^x x^{k} dx = \int_{0}^{\infty} x^{k} e^{\ln(a)x} dx= \int_{0}^{\infty} x^{k} e^{-\ln(\frac{1}{a})x} dx = \frac{\Gamma(k+1)}{(-\ln(a))^{k+1}}. $$

The last integral can be evaluate by using the change of variables $ t=\ln\left (\frac{1}{a} \right)x $ and the gamma function

$$ \Gamma( s+1 ) = \int_{0}^{\infty} t^{s}e^{-t}dt. $$

Note: Here are some numerical values for the sum $S$ and the integral $I$ which show they are almost equal

$$ (k,a)=(3,1/2) \implies S = 26.00000000,\quad I = 25.99258100 $$

$$ (k,a)=(10,1/7) \implies S = 2395.590764,\quad I = 2395.589595 $$

Answer 2

Let: \begin{align} a&\in (0,1)\\\ k&\in\mathbb{N}\\ f_k&\equiv \sum_{n=1}^\infty a^n n^k\\\ g_k&\equiv \int_{x=0}^\infty a^x x^k dx. \end{align} Then: \begin{align} f_0 &= \frac{a}{1-a}\\\ \\ f_k &= a\frac{df_{k-1}}{da}\\\ \\ g_k &= \frac{(-1)^{k+1}k!}{\ln(a)^{k+1}}. \end{align}

For $g_k$ we're lucky enough to have an explicit solution (as pointed out by Mhenni). The recurrence relation for $f_k$ together with the initial condition uniquely define the entire series $f_k\ \forall\ k\in\mathbb{N}$: \begin{align} f_1 &= \left(\frac{a}{1-a}\right) + \left(\frac{a}{1-a}\right)^2 = \frac{a}{(a-1)^2}\\ f_2 &= \left(\frac{a}{1-a}\right) + 3\left(\frac{a}{1-a}\right)^2 + 2\left(\frac{a}{1-a}\right)^3 = \frac{a(a+1)}{(1-a)^3}\\ f_3 &= \left(\frac{a}{1-a}\right) + 7\left(\frac{a}{1-a}\right)^2 + 12\left(\frac{a}{1-a}\right)^3 + 6\left(\frac{a}{1-a}\right)^4=\frac{a(a^2+1+4a)}{(a-1)^4}\\ &\ldots \end{align} Each of the terms in the $f_k$ are positive, so $f_{k+1}>f_k\ \forall\ k\in \mathbb{N}$.

Using the above, one can define the ratio $C_k \equiv f_k/g_k$. All of the various $C_k$ are finite functions of $a$, which vary smoothly in the interval $a\in(0,1)$. However, if you want to define the inequality $f_k\leq Cg_k\ \forall k\in\mathbb{N}\ \forall a\in(0,1)$ for some $C\in \mathbb{R}$ you would need $C=\max(C_k)$ where the maximization is over both $k$ and $a$. I'm not sure if such a $C$ exists or not.

Answer 3

Given $n,k\in\mathbb{N}$ and $a\in(0,1)$, when $x\in[n,n+1]$, $a^xx^k>a^{n+1}n^k$. It follows that $$\int_0^\infty a^xx^kdx>\sum_{n=1}^\infty\int_n^{n+1}a^xx^kdx>\sum_{n=1}^\infty a^{n+1}n^k=a\cdot\sum_{n=1}^\infty a^nn^k.$$

Edit: The argument above shows that $C(a)$ can be chosen as $a^{-1}$. Actually $C$ cannot be independent of $a$. From $\sum_{n=1}^\infty a^nn^k>a$ and $\int_0^\infty a^x x^kdx=\frac{k!}{(\log a^{-1})^{k+1}}$ we know that $$R(a,k):=\frac{\sum_{n=1}^\infty a^nn^k}{\int_0^\infty a^x x^kdx}>\frac{a\cdot(\log a^{-1})^{k+1}}{k!}.$$ Let $a=e^{-k}$, then $$R(e^{-k},k)>\frac{k^{k+1}}{e^k\cdot k!}.$$ Due to Stirling' formula, $\lim_{k\to\infty}R(e^{-k},k)=\infty$.