Prove that $f(x) = sum_{n=1}^{infty} x^n/n^2$ is continuous on $[0,1]$












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I have a general idea of how to prove this but I could use some help with the details.



Basically I see that $f(x)$ is the uniform limit of $f_k(x) = sum_{n=1}^{k} x^n/n^2$ on $[0,1]$.



Each $f_k$ is continuous, so $f$ is as well since uniform convergence preserves continuity.



Is this proof correct/does it seem sufficient?



Thanks ahead of time.






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    The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 4:53
















5












$begingroup$


I have a general idea of how to prove this but I could use some help with the details.



Basically I see that $f(x)$ is the uniform limit of $f_k(x) = sum_{n=1}^{k} x^n/n^2$ on $[0,1]$.



Each $f_k$ is continuous, so $f$ is as well since uniform convergence preserves continuity.



Is this proof correct/does it seem sufficient?



Thanks ahead of time.






real-analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 4:53














5












5








5


2



$begingroup$


I have a general idea of how to prove this but I could use some help with the details.



Basically I see that $f(x)$ is the uniform limit of $f_k(x) = sum_{n=1}^{k} x^n/n^2$ on $[0,1]$.



Each $f_k$ is continuous, so $f$ is as well since uniform convergence preserves continuity.



Is this proof correct/does it seem sufficient?



Thanks ahead of time.






real-analysis proof-verification proof-writing







share|cite|improve this question











$endgroup$




I have a general idea of how to prove this but I could use some help with the details.



Basically I see that $f(x)$ is the uniform limit of $f_k(x) = sum_{n=1}^{k} x^n/n^2$ on $[0,1]$.



Each $f_k$ is continuous, so $f$ is as well since uniform convergence preserves continuity.



Is this proof correct/does it seem sufficient?



Thanks ahead of time.






real-analysis proof-verification proof-writing




real-analysis proof-verification proof-writing






share|cite|improve this question















share|cite|improve this question













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edited Nov 30 '18 at 4:53







user591271

















asked Nov 30 '18 at 4:49









user591271user591271

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  • 2




    $begingroup$
    The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 4:53














  • 2




    $begingroup$
    The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 4:53








2




2




$begingroup$
The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here.
$endgroup$
– Lord Shark the Unknown
Nov 30 '18 at 4:53




$begingroup$
The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here.
$endgroup$
– Lord Shark the Unknown
Nov 30 '18 at 4:53










2 Answers
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Note that $|x^n/n^2|leq 1/n^2$ on $[0,1]$ and $sum_{n=1}^{infty} 1/n^2 lt infty$. Then by Weierstrass M-test,$sum_{n=1}^{infty} x^n/n^2$ converges uniformly. Hence $f$ is continuous.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    This can be done by a calculus student, actually, and I can even prove the continuity is uniform, provided said calculus student recalls the mean value theorem. No $M$-test needed here!



    Let $x_0, x_1 in [0,1]$, and let $epsilon > 0$ be given. We'll find $delta$ so that $|x_0 - x_1| < delta implies |f(x_0) - f(x_1)| < epsilon$.



    To see this, introduce the auxiliary functions $g_n(x) = x^n$



    We have:



    $$|f(x_1) - f(x_0)| = bigg|sum_{n=1}^infty frac{x_1^n - x_0^n}{n^2} bigg| = bigg| sum_{n=1}^infty frac{g_n(x_1) - g_n(x_0)}{n^2}bigg|$$



    Now $g_n$ are all differentiable, so $g_n(x_1) - g_n(x_0) = g_n'(x)(x_1 - x_0)$ by the Mean Value Theorem, for some $x in (x_0, x_1)$ (assuming without loss of generality $x_1 > x_0$.



    Now we have $g_n'(x) = nx^{n-1}$, so shoving this in to the above equation and observing that now all the terms are positive in the sum, so we drop the bars:



    $$|f(x_1) - f(x_0)| leq bigg[sum_{n=1}^infty frac{x^{n-1}}{n}bigg] |x_1 -x_0|$$



    But now, the first term is a convergent series, since $x < 1$, and $n > 0$, this expression is bounded above by a geometric series. Calling the limit value $S$, taking $delta = 1/S$ suffices.






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      2 Answers
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      5












      $begingroup$

      Note that $|x^n/n^2|leq 1/n^2$ on $[0,1]$ and $sum_{n=1}^{infty} 1/n^2 lt infty$. Then by Weierstrass M-test,$sum_{n=1}^{infty} x^n/n^2$ converges uniformly. Hence $f$ is continuous.






      share|cite|improve this answer











      $endgroup$


















        5












        $begingroup$

        Note that $|x^n/n^2|leq 1/n^2$ on $[0,1]$ and $sum_{n=1}^{infty} 1/n^2 lt infty$. Then by Weierstrass M-test,$sum_{n=1}^{infty} x^n/n^2$ converges uniformly. Hence $f$ is continuous.






        share|cite|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          Note that $|x^n/n^2|leq 1/n^2$ on $[0,1]$ and $sum_{n=1}^{infty} 1/n^2 lt infty$. Then by Weierstrass M-test,$sum_{n=1}^{infty} x^n/n^2$ converges uniformly. Hence $f$ is continuous.






          share|cite|improve this answer











          $endgroup$



          Note that $|x^n/n^2|leq 1/n^2$ on $[0,1]$ and $sum_{n=1}^{infty} 1/n^2 lt infty$. Then by Weierstrass M-test,$sum_{n=1}^{infty} x^n/n^2$ converges uniformly. Hence $f$ is continuous.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '18 at 5:10

























          answered Nov 30 '18 at 5:05









          Thomas ShelbyThomas Shelby

          2,260220




          2,260220























              1












              $begingroup$

              This can be done by a calculus student, actually, and I can even prove the continuity is uniform, provided said calculus student recalls the mean value theorem. No $M$-test needed here!



              Let $x_0, x_1 in [0,1]$, and let $epsilon > 0$ be given. We'll find $delta$ so that $|x_0 - x_1| < delta implies |f(x_0) - f(x_1)| < epsilon$.



              To see this, introduce the auxiliary functions $g_n(x) = x^n$



              We have:



              $$|f(x_1) - f(x_0)| = bigg|sum_{n=1}^infty frac{x_1^n - x_0^n}{n^2} bigg| = bigg| sum_{n=1}^infty frac{g_n(x_1) - g_n(x_0)}{n^2}bigg|$$



              Now $g_n$ are all differentiable, so $g_n(x_1) - g_n(x_0) = g_n'(x)(x_1 - x_0)$ by the Mean Value Theorem, for some $x in (x_0, x_1)$ (assuming without loss of generality $x_1 > x_0$.



              Now we have $g_n'(x) = nx^{n-1}$, so shoving this in to the above equation and observing that now all the terms are positive in the sum, so we drop the bars:



              $$|f(x_1) - f(x_0)| leq bigg[sum_{n=1}^infty frac{x^{n-1}}{n}bigg] |x_1 -x_0|$$



              But now, the first term is a convergent series, since $x < 1$, and $n > 0$, this expression is bounded above by a geometric series. Calling the limit value $S$, taking $delta = 1/S$ suffices.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                This can be done by a calculus student, actually, and I can even prove the continuity is uniform, provided said calculus student recalls the mean value theorem. No $M$-test needed here!



                Let $x_0, x_1 in [0,1]$, and let $epsilon > 0$ be given. We'll find $delta$ so that $|x_0 - x_1| < delta implies |f(x_0) - f(x_1)| < epsilon$.



                To see this, introduce the auxiliary functions $g_n(x) = x^n$



                We have:



                $$|f(x_1) - f(x_0)| = bigg|sum_{n=1}^infty frac{x_1^n - x_0^n}{n^2} bigg| = bigg| sum_{n=1}^infty frac{g_n(x_1) - g_n(x_0)}{n^2}bigg|$$



                Now $g_n$ are all differentiable, so $g_n(x_1) - g_n(x_0) = g_n'(x)(x_1 - x_0)$ by the Mean Value Theorem, for some $x in (x_0, x_1)$ (assuming without loss of generality $x_1 > x_0$.



                Now we have $g_n'(x) = nx^{n-1}$, so shoving this in to the above equation and observing that now all the terms are positive in the sum, so we drop the bars:



                $$|f(x_1) - f(x_0)| leq bigg[sum_{n=1}^infty frac{x^{n-1}}{n}bigg] |x_1 -x_0|$$



                But now, the first term is a convergent series, since $x < 1$, and $n > 0$, this expression is bounded above by a geometric series. Calling the limit value $S$, taking $delta = 1/S$ suffices.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  This can be done by a calculus student, actually, and I can even prove the continuity is uniform, provided said calculus student recalls the mean value theorem. No $M$-test needed here!



                  Let $x_0, x_1 in [0,1]$, and let $epsilon > 0$ be given. We'll find $delta$ so that $|x_0 - x_1| < delta implies |f(x_0) - f(x_1)| < epsilon$.



                  To see this, introduce the auxiliary functions $g_n(x) = x^n$



                  We have:



                  $$|f(x_1) - f(x_0)| = bigg|sum_{n=1}^infty frac{x_1^n - x_0^n}{n^2} bigg| = bigg| sum_{n=1}^infty frac{g_n(x_1) - g_n(x_0)}{n^2}bigg|$$



                  Now $g_n$ are all differentiable, so $g_n(x_1) - g_n(x_0) = g_n'(x)(x_1 - x_0)$ by the Mean Value Theorem, for some $x in (x_0, x_1)$ (assuming without loss of generality $x_1 > x_0$.



                  Now we have $g_n'(x) = nx^{n-1}$, so shoving this in to the above equation and observing that now all the terms are positive in the sum, so we drop the bars:



                  $$|f(x_1) - f(x_0)| leq bigg[sum_{n=1}^infty frac{x^{n-1}}{n}bigg] |x_1 -x_0|$$



                  But now, the first term is a convergent series, since $x < 1$, and $n > 0$, this expression is bounded above by a geometric series. Calling the limit value $S$, taking $delta = 1/S$ suffices.






                  share|cite|improve this answer











                  $endgroup$



                  This can be done by a calculus student, actually, and I can even prove the continuity is uniform, provided said calculus student recalls the mean value theorem. No $M$-test needed here!



                  Let $x_0, x_1 in [0,1]$, and let $epsilon > 0$ be given. We'll find $delta$ so that $|x_0 - x_1| < delta implies |f(x_0) - f(x_1)| < epsilon$.



                  To see this, introduce the auxiliary functions $g_n(x) = x^n$



                  We have:



                  $$|f(x_1) - f(x_0)| = bigg|sum_{n=1}^infty frac{x_1^n - x_0^n}{n^2} bigg| = bigg| sum_{n=1}^infty frac{g_n(x_1) - g_n(x_0)}{n^2}bigg|$$



                  Now $g_n$ are all differentiable, so $g_n(x_1) - g_n(x_0) = g_n'(x)(x_1 - x_0)$ by the Mean Value Theorem, for some $x in (x_0, x_1)$ (assuming without loss of generality $x_1 > x_0$.



                  Now we have $g_n'(x) = nx^{n-1}$, so shoving this in to the above equation and observing that now all the terms are positive in the sum, so we drop the bars:



                  $$|f(x_1) - f(x_0)| leq bigg[sum_{n=1}^infty frac{x^{n-1}}{n}bigg] |x_1 -x_0|$$



                  But now, the first term is a convergent series, since $x < 1$, and $n > 0$, this expression is bounded above by a geometric series. Calling the limit value $S$, taking $delta = 1/S$ suffices.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 6:54

























                  answered Nov 30 '18 at 6:44









                  Alfred YergerAlfred Yerger

                  10.3k2148




                  10.3k2148






























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