Proof of radius of convergence for geometric series.












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I am just getting into proofs involving a radius of convergence and I am tasked with proving for every $b$ with $0<b<1$ the series $sum_{n=0}^infty x^n$ converges uniformly to $frac{1}{1-x}$ on the interval $[-b,b]$ while working directly with the definition of uniform convergence.
So far I have $lvert f(x) - s_k (x) rvert = lvert frac{1}{1-x}-frac{1-x^{k+1}}{1-x}rvert = lvert frac{x^{k+1}}{1-x}rvert < epsilon$ But now I am stuck as far choosing an $N$ and doing the rest of the proof. If anyone could help me that would be greatly appriciated thank you!






real-analysis power-series uniform-convergence







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    $begingroup$


    I am just getting into proofs involving a radius of convergence and I am tasked with proving for every $b$ with $0<b<1$ the series $sum_{n=0}^infty x^n$ converges uniformly to $frac{1}{1-x}$ on the interval $[-b,b]$ while working directly with the definition of uniform convergence.
    So far I have $lvert f(x) - s_k (x) rvert = lvert frac{1}{1-x}-frac{1-x^{k+1}}{1-x}rvert = lvert frac{x^{k+1}}{1-x}rvert < epsilon$ But now I am stuck as far choosing an $N$ and doing the rest of the proof. If anyone could help me that would be greatly appriciated thank you!






    real-analysis power-series uniform-convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am just getting into proofs involving a radius of convergence and I am tasked with proving for every $b$ with $0<b<1$ the series $sum_{n=0}^infty x^n$ converges uniformly to $frac{1}{1-x}$ on the interval $[-b,b]$ while working directly with the definition of uniform convergence.
      So far I have $lvert f(x) - s_k (x) rvert = lvert frac{1}{1-x}-frac{1-x^{k+1}}{1-x}rvert = lvert frac{x^{k+1}}{1-x}rvert < epsilon$ But now I am stuck as far choosing an $N$ and doing the rest of the proof. If anyone could help me that would be greatly appriciated thank you!






      real-analysis power-series uniform-convergence







      share|cite|improve this question











      $endgroup$




      I am just getting into proofs involving a radius of convergence and I am tasked with proving for every $b$ with $0<b<1$ the series $sum_{n=0}^infty x^n$ converges uniformly to $frac{1}{1-x}$ on the interval $[-b,b]$ while working directly with the definition of uniform convergence.
      So far I have $lvert f(x) - s_k (x) rvert = lvert frac{1}{1-x}-frac{1-x^{k+1}}{1-x}rvert = lvert frac{x^{k+1}}{1-x}rvert < epsilon$ But now I am stuck as far choosing an $N$ and doing the rest of the proof. If anyone could help me that would be greatly appriciated thank you!






      real-analysis power-series uniform-convergence




      real-analysis power-series uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 25 at 19:08









      José Carlos Santos

      169k23132237




      169k23132237










      asked Jan 25 at 19:03









      MosephMoseph

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          $begingroup$

          Since, $xin[-b,b]$,$$leftlvertfrac{x^{k+1}}{1-x}rightrvert=frac{lvert xrvert^{k+1}}{lvert1-xrvert}leqslantfrac{lvert brvert^{k+1}}{1-lvert brvert}.tag1$$Since $lvert brvert<1$,$$lim_{ktoinfty}frac{lvert brvert^{k+1}}{1-lvert brvert}=0.$$So, given $varepsilon>0$, there is a natural $N$ such that$$kgeqslant Nimpliesfrac{lvert brvert^{k+1}}{1-lvert brvert}<varepsilon$$and it now follows from $(1)$ that$$kgeqslant Nimpliesbigl(forall xin[-b,b]bigr):leftlvertfrac{x^{k+1}}{1-x}rightrvert<varepsilon.$$






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          • $begingroup$
            I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
            $endgroup$
            – Moseph
            Jan 25 at 19:37












          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Jan 25 at 19:41











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          $begingroup$

          Since, $xin[-b,b]$,$$leftlvertfrac{x^{k+1}}{1-x}rightrvert=frac{lvert xrvert^{k+1}}{lvert1-xrvert}leqslantfrac{lvert brvert^{k+1}}{1-lvert brvert}.tag1$$Since $lvert brvert<1$,$$lim_{ktoinfty}frac{lvert brvert^{k+1}}{1-lvert brvert}=0.$$So, given $varepsilon>0$, there is a natural $N$ such that$$kgeqslant Nimpliesfrac{lvert brvert^{k+1}}{1-lvert brvert}<varepsilon$$and it now follows from $(1)$ that$$kgeqslant Nimpliesbigl(forall xin[-b,b]bigr):leftlvertfrac{x^{k+1}}{1-x}rightrvert<varepsilon.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
            $endgroup$
            – Moseph
            Jan 25 at 19:37












          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Jan 25 at 19:41
















          2












          $begingroup$

          Since, $xin[-b,b]$,$$leftlvertfrac{x^{k+1}}{1-x}rightrvert=frac{lvert xrvert^{k+1}}{lvert1-xrvert}leqslantfrac{lvert brvert^{k+1}}{1-lvert brvert}.tag1$$Since $lvert brvert<1$,$$lim_{ktoinfty}frac{lvert brvert^{k+1}}{1-lvert brvert}=0.$$So, given $varepsilon>0$, there is a natural $N$ such that$$kgeqslant Nimpliesfrac{lvert brvert^{k+1}}{1-lvert brvert}<varepsilon$$and it now follows from $(1)$ that$$kgeqslant Nimpliesbigl(forall xin[-b,b]bigr):leftlvertfrac{x^{k+1}}{1-x}rightrvert<varepsilon.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
            $endgroup$
            – Moseph
            Jan 25 at 19:37












          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Jan 25 at 19:41














          2












          2








          2





          $begingroup$

          Since, $xin[-b,b]$,$$leftlvertfrac{x^{k+1}}{1-x}rightrvert=frac{lvert xrvert^{k+1}}{lvert1-xrvert}leqslantfrac{lvert brvert^{k+1}}{1-lvert brvert}.tag1$$Since $lvert brvert<1$,$$lim_{ktoinfty}frac{lvert brvert^{k+1}}{1-lvert brvert}=0.$$So, given $varepsilon>0$, there is a natural $N$ such that$$kgeqslant Nimpliesfrac{lvert brvert^{k+1}}{1-lvert brvert}<varepsilon$$and it now follows from $(1)$ that$$kgeqslant Nimpliesbigl(forall xin[-b,b]bigr):leftlvertfrac{x^{k+1}}{1-x}rightrvert<varepsilon.$$






          share|cite|improve this answer











          $endgroup$



          Since, $xin[-b,b]$,$$leftlvertfrac{x^{k+1}}{1-x}rightrvert=frac{lvert xrvert^{k+1}}{lvert1-xrvert}leqslantfrac{lvert brvert^{k+1}}{1-lvert brvert}.tag1$$Since $lvert brvert<1$,$$lim_{ktoinfty}frac{lvert brvert^{k+1}}{1-lvert brvert}=0.$$So, given $varepsilon>0$, there is a natural $N$ such that$$kgeqslant Nimpliesfrac{lvert brvert^{k+1}}{1-lvert brvert}<varepsilon$$and it now follows from $(1)$ that$$kgeqslant Nimpliesbigl(forall xin[-b,b]bigr):leftlvertfrac{x^{k+1}}{1-x}rightrvert<varepsilon.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 25 at 19:40

























          answered Jan 25 at 19:07









          José Carlos SantosJosé Carlos Santos

          169k23132237




          169k23132237












          • $begingroup$
            I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
            $endgroup$
            – Moseph
            Jan 25 at 19:37












          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Jan 25 at 19:41


















          • $begingroup$
            I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
            $endgroup$
            – Moseph
            Jan 25 at 19:37












          • $begingroup$
            I've edited my answer. What do you think now?
            $endgroup$
            – José Carlos Santos
            Jan 25 at 19:41
















          $begingroup$
          I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
          $endgroup$
          – Moseph
          Jan 25 at 19:37






          $begingroup$
          I also saw that but I am still confused on where to go from there. The definition of uniform convergence I am given is: Let $f_n (x)$ be a sequence of functions defined on a set $A$ in the reals. Then $f_n (x)$ converges uniformly on $A$ to limit function $f$ defined on $A$ if, for every $epsilon > 0$ there exists an $N$ in the natural numbers such that $lvert f_n (x) - f(x) rvert < epsilon$ whenever $ngeq N$ and $ x in A$
          $endgroup$
          – Moseph
          Jan 25 at 19:37














          $begingroup$
          I've edited my answer. What do you think now?
          $endgroup$
          – José Carlos Santos
          Jan 25 at 19:41




          $begingroup$
          I've edited my answer. What do you think now?
          $endgroup$
          – José Carlos Santos
          Jan 25 at 19:41


















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