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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.

vbm
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  • See https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6 – lab bhattacharjee Mar 06 '18 at 11:08
  • A lengthy process. How to show $M_n$ for the series. – vbm Mar 06 '18 at 11:47
  • Showing the series is uniformly convergent on $[-a,a]$ is much easier than evaluating the partial sums. Use the Weierstrass $M$-test, $|n^p x^n| \le n^p a^n$, and the root test. – GEdgar Mar 06 '18 at 11:59
  • @GEdgar, would you like to clear my two confusions? Is it everytime necessary that the $M_n$ , determined from a series must be of $\frac{1}{n^p}, p>1$ form? and following your suggestion if $|n^px^n| \leq n^pa^n$ and then if we take root test it won't be of the form $\sum\frac{1}{n^p}$. Is $n^pa^n$ correct to consider as $M_n$ – vbm Mar 06 '18 at 19:37

2 Answers2

1

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.

user
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  • If the numerator part is bounded then Abel's theorem can be applied. – vbm Mar 06 '18 at 13:31
  • @thevbm thelord Probably I misunderstood your question. My answer was not intended for proof of convergence. It is the result of summation in the convergence domain $|x|<1$. – user Mar 06 '18 at 13:55
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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 $[-a,a]$.

GEdgar
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  • I did the same thing but was still confused. Thank you for your great effort. It is now clear to me. – vbm Mar 07 '18 at 07:50