Finding where a series converges pointwise and uniformly












1












$begingroup$


Consider



$$sum_{n=0}^{infty} x(1 - x)^{2n} $$



Where does the sum converge pointwise and uniformly?



I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous






real-analysis sequences-and-series







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  • 5




    $begingroup$
    What has the title to do with the question?
    $endgroup$
    – egreg
    Dec 14 '18 at 21:08
















1












$begingroup$


Consider



$$sum_{n=0}^{infty} x(1 - x)^{2n} $$



Where does the sum converge pointwise and uniformly?



I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    What has the title to do with the question?
    $endgroup$
    – egreg
    Dec 14 '18 at 21:08














1












1








1





$begingroup$


Consider



$$sum_{n=0}^{infty} x(1 - x)^{2n} $$



Where does the sum converge pointwise and uniformly?



I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




Consider



$$sum_{n=0}^{infty} x(1 - x)^{2n} $$



Where does the sum converge pointwise and uniformly?



I think $[0, 2)$ pointwise and $(0, 2)$ uniformly because it becomes a geometric series. Also the function to which it converges to needs to be continuous






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 7:06









RRL

52.3k42573




52.3k42573










asked Dec 14 '18 at 21:05









josephjoseph

496111




496111








  • 5




    $begingroup$
    What has the title to do with the question?
    $endgroup$
    – egreg
    Dec 14 '18 at 21:08














  • 5




    $begingroup$
    What has the title to do with the question?
    $endgroup$
    – egreg
    Dec 14 '18 at 21:08








5




5




$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08




$begingroup$
What has the title to do with the question?
$endgroup$
– egreg
Dec 14 '18 at 21:08










1 Answer
1






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1












$begingroup$

You are correct about recognizing the geometric series converging pointwise on $[0,2)$.



For this series to converge uniformly on $(0,2)$ it is necessary that



$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$



However,



$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$



and



$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$



So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So it's uniform only at $0$?
    $endgroup$
    – joseph
    Dec 14 '18 at 23:56










  • $begingroup$
    @joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
    $endgroup$
    – RRL
    Dec 15 '18 at 0:10













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1 Answer
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1 Answer
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1












$begingroup$

You are correct about recognizing the geometric series converging pointwise on $[0,2)$.



For this series to converge uniformly on $(0,2)$ it is necessary that



$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$



However,



$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$



and



$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$



So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So it's uniform only at $0$?
    $endgroup$
    – joseph
    Dec 14 '18 at 23:56










  • $begingroup$
    @joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
    $endgroup$
    – RRL
    Dec 15 '18 at 0:10


















1












$begingroup$

You are correct about recognizing the geometric series converging pointwise on $[0,2)$.



For this series to converge uniformly on $(0,2)$ it is necessary that



$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$



However,



$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$



and



$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$



So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So it's uniform only at $0$?
    $endgroup$
    – joseph
    Dec 14 '18 at 23:56










  • $begingroup$
    @joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
    $endgroup$
    – RRL
    Dec 15 '18 at 0:10
















1












1








1





$begingroup$

You are correct about recognizing the geometric series converging pointwise on $[0,2)$.



For this series to converge uniformly on $(0,2)$ it is necessary that



$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$



However,



$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$



and



$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$



So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.






share|cite|improve this answer











$endgroup$



You are correct about recognizing the geometric series converging pointwise on $[0,2)$.



For this series to converge uniformly on $(0,2)$ it is necessary that



$$tag{*}sup_{x in (0,2)}left|sum_{n=N}^infty x(1-x)^{2n}right| xrightarrow[N to infty]{} 0$$



However,



$$left|sum_{n=N}^infty x(1-x)^{2n}right| = x(1-x)^{2N}sum_{n=0}^infty (1-x)^{2n} = frac{x(1-x)^{2N}}{1 - (1-x)^2} = frac{(1-x)^{2N}}{2 - x},$$



and



$$sup_{x in (0,2)}frac{(1-x)^{2N}}{2 - x} = +infty$$



So the convergence is not uniform on $(0,2)$. Noting that a problem arises due to behavior near $x = 2$ (and $x=0$), try on your own to determine and justify where convergence is uniform.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 27 '18 at 6:32

























answered Dec 14 '18 at 21:50









RRLRRL

52.3k42573




52.3k42573












  • $begingroup$
    So it's uniform only at $0$?
    $endgroup$
    – joseph
    Dec 14 '18 at 23:56










  • $begingroup$
    @joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
    $endgroup$
    – RRL
    Dec 15 '18 at 0:10




















  • $begingroup$
    So it's uniform only at $0$?
    $endgroup$
    – joseph
    Dec 14 '18 at 23:56










  • $begingroup$
    @joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
    $endgroup$
    – RRL
    Dec 15 '18 at 0:10


















$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56




$begingroup$
So it's uniform only at $0$?
$endgroup$
– joseph
Dec 14 '18 at 23:56












$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10






$begingroup$
@joseph: No -- uniform convergence is relevant for a set of values. So we say the series $sum{n=1}^infty f_n(x)$ converges uniformly to $S(x)$ for $x in (a,b)$ if it converges pointwise and $|sum_{n=1}^N f_n(x) - S(x)| < epsilon$ when $N > N_0$ for ALL $x in (a,b)$ where $N_0$ does NOT depend on $x$. I just showed your series is not UC on $(0,2)$ but what about $(0,b]$ where $b < 2$?
$endgroup$
– RRL
Dec 15 '18 at 0:10




















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