Show that the series $sumlimits_{k=1}^{infty}frac{x^k}{k}$ does not converge uniformly on $(-1,1)$.
$begingroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
$endgroup$
add a comment |
$begingroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
$endgroup$
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
add a comment |
$begingroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
$endgroup$
Show that the series $sumlimits_{k=1}^{infty}dfrac{x^k}{k}$ does not converge uniformly on $(-1,1)$. (Hint: Show that the sequence of partial sums is not Cauchy in the sup-norm.)
Proof: Let $s_n = sumlimits_{k=1}^{n}dfrac{x^k}{k}$. We need to show that there exists $epsilon>0$ such that for all $Ninmathbb{N}$ and for $n geq N$ implies $|s_n-s_N|_{sup} geq epsilon$. As a matter of fact, I claim that this must be true for any $epsilon >0$ we choose. So let $epsilon >0$. But,
begin{align}
|s_n-s_N|_{sup} &= |sumlimits_{k=1}^{n}dfrac{x^k}{k} - sumlimits_{k=1}^{N}dfrac{x^k}{k}|_{sup} \
&= |sumlimits_{k=N+1}^{n}dfrac{x^k}{k}|_{sup} \
&= sumlimits_{k=N+1}^{n}dfrac{1}{k}
end{align}
which is a harmonic series and will diverge to infinity as $ntoinfty$. $blacksquare$
real-analysis sequences-and-series proof-verification uniform-convergence
real-analysis sequences-and-series proof-verification uniform-convergence
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
edited Jan 2 at 21:44
Lorenzo B.
1,8602520
1,8602520
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
asked Apr 21 '17 at 19:23
user3000482user3000482
766518
766518
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
add a comment |
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40
add a comment |
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$begingroup$
Your proof is fine.
$endgroup$
– Starfall
Apr 21 '17 at 19:25
$begingroup$
Possible duplicate of A dubious proof using Weierstrass-M test for $sum^n_{k=1}frac{x^k}{k}$
$endgroup$
– Nosrati
Apr 21 '17 at 19:28
$begingroup$
you could calculate the value of series explicitly by integrating the simple geometric series ... then you will see that it is unbounded on $(-1,1)$.
$endgroup$
– Red shoes
Apr 21 '17 at 19:40