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Prove that $sum_n=0^inftyfracx^nn!$ is not uniformly convergent on $(0, +infty)ni x$.

prove that the series is not uniformly convergentProve $x^n$ is not uniformly convergentProve that function is not uniformly convergentProve that $f_n(x) = frac1x chi _[frac 1 n, 2]$ is not uniformly convergent on $[0,2]$?Prove $sum_n=0^inftyx^n$ not converge uniformlyProve that $sum_n=1^infty (fracln xx)^n$ converge uniformlyprove that $lim_nto infty a_n=$finiteProve convergence / divergence of $sum_n=2^infty(-1)^nfrac sqrt n(-1)^n+sqrt nsinleft(frac 1sqrt nright)$Prove that $sum_ngeqslant 1 x^n/n$ is not uniformly convergentWhy $ sum_k=1^inftyfrac1k^1+x $ is not uniformly convergent on $ left(0,;+inftyright) $

1

$begingroup$

Prove that $sum_n=0^inftyfracx^nn!$ is not uniformly
convergent on $(0, +infty)ni x$.

I wanted to do it by applying Cauchy's test and coming to a contradiction (i.e. $epsilon>$ positive constant):

$$epsilon>|fracx^n+1(n+1)!+fracx^n+2(n+2)!+cdots+fracx^n+p(n+p)!|$$

$$epsilon>|fracx^n+1(n+1)!cdot(1+fracxn+2+cdots+fracx^p-1(n+2)cdots(n+p))|.$$

The expressions within the ||'s reminded me of $e^x=1+x+cdots$, but after several tries I still wasn't able to evaluate. I've also tried standard method of 'find the smallest one in the brackets and take it $p$ times', but that didn't lead anywhere.

Another idea where I took $x$ to be an expression involving $n$ worked, but is it even legal? $x$ is supposed to be fixed, and $n$ is arbitrarily large, so I don't know.

Hints or full answers, I will be very grateful.

sequences-and-series convergence uniform-convergence

share|cite|improve this question

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could use the fact that a uniformly convergent series is uniformly Cauchy. In particular, its terms must converge uniformly to $0$.
  $endgroup$
  – David Mitra
  Apr 1 at 17:05
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Prove that $sum_n=0^inftyfracx^nn!$ is not uniformly
convergent on $(0, +infty)ni x$.

I wanted to do it by applying Cauchy's test and coming to a contradiction (i.e. $epsilon>$ positive constant):

$$epsilon>|fracx^n+1(n+1)!+fracx^n+2(n+2)!+cdots+fracx^n+p(n+p)!|$$

$$epsilon>|fracx^n+1(n+1)!cdot(1+fracxn+2+cdots+fracx^p-1(n+2)cdots(n+p))|.$$

The expressions within the ||'s reminded me of $e^x=1+x+cdots$, but after several tries I still wasn't able to evaluate. I've also tried standard method of 'find the smallest one in the brackets and take it $p$ times', but that didn't lead anywhere.

Another idea where I took $x$ to be an expression involving $n$ worked, but is it even legal? $x$ is supposed to be fixed, and $n$ is arbitrarily large, so I don't know.

Hints or full answers, I will be very grateful.

sequences-and-series convergence uniform-convergence

share|cite|improve this question

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could use the fact that a uniformly convergent series is uniformly Cauchy. In particular, its terms must converge uniformly to $0$.
  $endgroup$
  – David Mitra
  Apr 1 at 17:05
  
  

  
  
  
  

add a comment | 

$begingroup$

Prove that $sum_n=0^inftyfracx^nn!$ is not uniformly
convergent on $(0, +infty)ni x$.

I wanted to do it by applying Cauchy's test and coming to a contradiction (i.e. $epsilon>$ positive constant):

$$epsilon>|fracx^n+1(n+1)!+fracx^n+2(n+2)!+cdots+fracx^n+p(n+p)!|$$

$$epsilon>|fracx^n+1(n+1)!cdot(1+fracxn+2+cdots+fracx^p-1(n+2)cdots(n+p))|.$$

The expressions within the ||'s reminded me of $e^x=1+x+cdots$, but after several tries I still wasn't able to evaluate. I've also tried standard method of 'find the smallest one in the brackets and take it $p$ times', but that didn't lead anywhere.

Another idea where I took $x$ to be an expression involving $n$ worked, but is it even legal? $x$ is supposed to be fixed, and $n$ is arbitrarily large, so I don't know.

Hints or full answers, I will be very grateful.

sequences-and-series convergence uniform-convergence

share|cite|improve this question

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

$endgroup$

share|cite|improve this question

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

share|cite|improve this question

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

asked Apr 1 at 16:47

fragileradiusfragileradius

306214

2 Answers
2

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oldest

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2

$begingroup$

Note that

$$sup_x in (0,infty)left|sum_k=n+1^infty fracx^kk!right| geqslantsup_x in (1,infty)sum_k=n+1^2n fracx^kk!geqslant sup_x in (1,infty)fracnx^n(2n)! $$

Can you show that the RHS does not converge to $0$ as $n to infty$?

share|cite|improve this answer

answered Apr 1 at 19:57

RRLRRL

53.4k52574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry, but I'm having trouble seeing how RHS does not converge to zero. Am I supposed to take $x=n$? Thank you.
  $endgroup$
  – fragileradius
  Apr 2 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well either see that the RHS is greater than $fracnx_n^n(2n)! $ for any sequence $x_n$ and take any $x_n^n$ growing faster than $(2n)!$ -- like $x_n = (2n)!$ -- or simply notice that the supremum is $+infty$.
  $endgroup$
  – RRL
  Apr 2 at 21:50
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Much easier if we can use that the sum is the exponential: the partial sum is a polynomial and the exponential grows faster than any polinomial, so for any $n$
$$sup_xin(0,infty)left|e^x - sum_k=0^nfracx^kk!right| = infty$$

share|cite|improve this answer

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

add a comment | 

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2

$begingroup$

Note that

$$sup_x in (0,infty)left|sum_k=n+1^infty fracx^kk!right| geqslantsup_x in (1,infty)sum_k=n+1^2n fracx^kk!geqslant sup_x in (1,infty)fracnx^n(2n)! $$

Can you show that the RHS does not converge to $0$ as $n to infty$?

share|cite|improve this answer

answered Apr 1 at 19:57

RRLRRL

53.4k52574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry, but I'm having trouble seeing how RHS does not converge to zero. Am I supposed to take $x=n$? Thank you.
  $endgroup$
  – fragileradius
  Apr 2 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well either see that the RHS is greater than $fracnx_n^n(2n)! $ for any sequence $x_n$ and take any $x_n^n$ growing faster than $(2n)!$ -- like $x_n = (2n)!$ -- or simply notice that the supremum is $+infty$.
  $endgroup$
  – RRL
  Apr 2 at 21:50
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that

$$sup_x in (0,infty)left|sum_k=n+1^infty fracx^kk!right| geqslantsup_x in (1,infty)sum_k=n+1^2n fracx^kk!geqslant sup_x in (1,infty)fracnx^n(2n)! $$

Can you show that the RHS does not converge to $0$ as $n to infty$?

share|cite|improve this answer

answered Apr 1 at 19:57

RRLRRL

53.4k52574

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry, but I'm having trouble seeing how RHS does not converge to zero. Am I supposed to take $x=n$? Thank you.
  $endgroup$
  – fragileradius
  Apr 2 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well either see that the RHS is greater than $fracnx_n^n(2n)! $ for any sequence $x_n$ and take any $x_n^n$ growing faster than $(2n)!$ -- like $x_n = (2n)!$ -- or simply notice that the supremum is $+infty$.
  $endgroup$
  – RRL
  Apr 2 at 21:50
  
  

  
  
  
  

add a comment | 

$begingroup$

Note that

$$sup_x in (0,infty)left|sum_k=n+1^infty fracx^kk!right| geqslantsup_x in (1,infty)sum_k=n+1^2n fracx^kk!geqslant sup_x in (1,infty)fracnx^n(2n)! $$

Can you show that the RHS does not converge to $0$ as $n to infty$?

share|cite|improve this answer

answered Apr 1 at 19:57

RRLRRL

53.4k52574

$endgroup$

share|cite|improve this answer

answered Apr 1 at 19:57

RRLRRL

53.4k52574

answered Apr 1 at 19:57

RRLRRL

53.4k52574

answered Apr 1 at 19:57

RRLRRL

53.4k52574

1

$begingroup$

Much easier if we can use that the sum is the exponential: the partial sum is a polynomial and the exponential grows faster than any polinomial, so for any $n$
$$sup_xin(0,infty)left|e^x - sum_k=0^nfracx^kk!right| = infty$$

share|cite|improve this answer

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

add a comment | 

1

$begingroup$

Much easier if we can use that the sum is the exponential: the partial sum is a polynomial and the exponential grows faster than any polinomial, so for any $n$
$$sup_xin(0,infty)left|e^x - sum_k=0^nfracx^kk!right| = infty$$

share|cite|improve this answer

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

add a comment | 

$begingroup$

Much easier if we can use that the sum is the exponential: the partial sum is a polynomial and the exponential grows faster than any polinomial, so for any $n$
$$sup_xin(0,infty)left|e^x - sum_k=0^nfracx^kk!right| = infty$$

share|cite|improve this answer

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

$endgroup$

share|cite|improve this answer

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972

answered Apr 2 at 16:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.4k42972