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Show that $\sum_{j=0}^{\infty} (j+1)x^j$ converges uniformly for x $\in$ any compact subset of (-1,1).

Using the ratio test, I got: $\frac{j+2}{j+1}x^j$. However, I don't know how to compare this to 1...Is there another method, or am I missing something?

kiwifruit
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    This series doesn't converge uniformly on $(-1,1)$; it converges uniformly on every compact subset of $(-1,1)$, though. –  Apr 27 '14 at 22:00
  • Ok, edited question – kiwifruit Apr 27 '14 at 22:02
  • Would compactness be equivalent to a closed interval? – kiwifruit Apr 27 '14 at 22:03
  • No, but compactness is equivalent to being a closed subset of $(-1,1)$. It would suffice to prove that the series converges uniformly on every interval of the form $[-\ell, \ell]$ for $0 \le \ell < 1$. –  Apr 27 '14 at 22:04
  • Ok, thank you...Still not sure how to use the convergence test results. It is clear that $x^j$ will be < 1, but I don't know about the fraction. – kiwifruit Apr 27 '14 at 22:06
  • Notice that $|x|^j \le \ell^j$ for all $x \in [-\ell, \ell]$, and try to combine it with the fact that $\frac{j + 2}{j + 1}$ is bounded. –  Apr 27 '14 at 22:07
  • Wouldn't the supremum of the fraction be something slightly greater than 1, so can that still guarantee the product is <1? – kiwifruit Apr 27 '14 at 22:09
  • You'll need to use the fact that $\ell^j \to 0$. –  Apr 27 '14 at 22:10
  • Hmm, but if we are using limsup, then wouldn't the supremum and hence its limit be something other than 0? – kiwifruit Apr 27 '14 at 22:12
  • No; the $\limsup$ is much smaller than the $\sup$. Try to compute $$\lim_{j \to \infty} \left(\frac{j + 2}{j + 1}\right) \ell^j$$ and convince yourself that it is, in fact, $0$. –  Apr 27 '14 at 22:14
  • Ok, I see that this is 0, but how can we take out the supremum part? Are we using the fact that the limit and limsup are equal when we have convergence? I think I am just confused...must the limsup eventually be equal to the limit? – kiwifruit Apr 27 '14 at 22:16
  • How is $\left(\frac{j+2}{j+1}\right)\ell^j$ related to the question? – Did Apr 27 '14 at 22:40
  • It is from the ratio test – kiwifruit Apr 27 '14 at 23:28

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Let $R$ in $(0,1)$. For every $x$ in $[-R,R]$ and every $n$, $$ \left|\sum_{j\geqslant n}(j+1)x^j\right|\leqslant\varepsilon_R(n), $$ where $$ \varepsilon_R(n)=\sum_{j\geqslant n}(j+1)R^j. $$ The question is to show that for every $R\lt1$, $$ \lim_{n\to\infty}\varepsilon_R(n)=0. $$ Can you show this?

Did
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