Which infinite series is this - $\sum_{n=1}^{\infty} n^{p}x^n$

Question

A Series is given $$f_n(x)=\sum_{n=1}^{\infty} n^{p}x^n$$, where $a \in (0,1), \forall p>0$. We have to show that the $f_n(x)$ is uniformly convergent on $[-a,a]$

From here, I get $$1^p x+2^p x^2+3^px^3+.....\infty$$ I know this series is uniformly convergent in the given interval where $\limsup_{n \to \infty}a^n \longrightarrow 0, x[-a,a]\subset (0,1)$

Now I'am stuck here. I cannot show $M_n$ from the series as I cannot reduce this in a particular format.

Anybody knows in which form of series this can be reduced? Any help is appreciated.

Answer 1

Proof of uniform convergence, without evaluating the partial sum.

Let $p>0$ and $0<a<1$. We claim the series $$ \sum_{n=1}^{\infty} n^{p}x^n \tag{1} $$ converges uniformly on $[-a,a]$.

Indeed, for any $x \in [-a,a]$, we have $$ |n^p x^n| \le n^p a^n . $$ The series $$ \sum_{n=1}^\infty n^p a^n $$ converges by the root test: as $n \to \infty$, $$ \big(n^p a^n\big)^{1/n} = (n^{1/n})^p \cdot a \to 1^p\cdot a = a < 1 . $$ Thus, by the Weierstrass M-test, series (1) converges uniformly on $[-a,a]$.

Answer 2

If $|x|<1$ the general expression for the sum is:

$$ \sum_{k=1}^\infty k^n x^k=\frac{x\sum_{l=0}^{n-1}A(n-1,l)x^l}{(1-x)^{n+1}}, $$ where $A(n-1,l)$ are the Eulerian numbers. Can be proved by induction.