Convergence of $sum_{n=1}^infty a_n$ when $a_n=f(frac 1 n)$ [duplicate]

1

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This question already has an answer here:

• 
  Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(frac{1}{i})$ (for large enough $i$)
  
  2 answers
  
  

I need help with this problem: 

Let $f$ be a continous function over an interval that contains $0$. Let $a_n=f(frac 1 n)$ (for $n$ large enough).

I've already showed that:

• If $sum_{n=1}^infty a_n$ converges, then $f(0)=0$.




• If $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.




• If $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.

How solve this one:

1. Suppose that $sum_{n=1}^infty a_n$ converges. Must f'(0) exist?


2. Suppose that $f(0)=f'(0)=0$. Should $sum_{n=1}^infty a_n$ converge?

calculus sequences-and-series convergence

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edited Mar 6 at 4:34

Robert Howard

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asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

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marked as duplicate by Clement C., Strants, Leucippus, YiFan, RRL Feb 2 at 6:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(frac{1}{i})$ (for large enough $i$)
  
  2 answers
  
  

I need help with this problem: 

Let $f$ be a continous function over an interval that contains $0$. Let $a_n=f(frac 1 n)$ (for $n$ large enough).

I've already showed that:

• If $sum_{n=1}^infty a_n$ converges, then $f(0)=0$.




• If $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.




• If $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.

How solve this one:

1. Suppose that $sum_{n=1}^infty a_n$ converges. Must f'(0) exist?


2. Suppose that $f(0)=f'(0)=0$. Should $sum_{n=1}^infty a_n$ converge?

calculus sequences-and-series convergence

share|cite|improve this question

edited Mar 6 at 4:34

Robert Howard

2,3033935

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

$endgroup$

marked as duplicate by Clement C., Strants, Leucippus, YiFan, RRL Feb 2 at 6:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

1

1

1

$begingroup$

This question already has an answer here:

• 
  Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(frac{1}{i})$ (for large enough $i$)
  
  2 answers
  
  

I need help with this problem: 

Let $f$ be a continous function over an interval that contains $0$. Let $a_n=f(frac 1 n)$ (for $n$ large enough).

I've already showed that:

• If $sum_{n=1}^infty a_n$ converges, then $f(0)=0$.




• If $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.




• If $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.

How solve this one:

1. Suppose that $sum_{n=1}^infty a_n$ converges. Must f'(0) exist?


2. Suppose that $f(0)=f'(0)=0$. Should $sum_{n=1}^infty a_n$ converge?

calculus sequences-and-series convergence

share|cite|improve this question

edited Mar 6 at 4:34

Robert Howard

2,3033935

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

$endgroup$

This question already has an answer here:

• 
  Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(frac{1}{i})$ (for large enough $i$)
  
  2 answers
  
  

I need help with this problem: 

Let $f$ be a continous function over an interval that contains $0$. Let $a_n=f(frac 1 n)$ (for $n$ large enough).

I've already showed that:

• If $sum_{n=1}^infty a_n$ converges, then $f(0)=0$.




• If $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.




• If $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.

How solve this one:

1. Suppose that $sum_{n=1}^infty a_n$ converges. Must f'(0) exist?


2. Suppose that $f(0)=f'(0)=0$. Should $sum_{n=1}^infty a_n$ converge?

This question already has an answer here:

• 
  Let $f$ be a continuous function on an interval around $0$ and let $a_i=f(frac{1}{i})$ (for large enough $i$)
  
  2 answers
  
  

calculus sequences-and-series convergence

calculus sequences-and-series convergence

share|cite|improve this question

edited Mar 6 at 4:34

Robert Howard

2,3033935

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

share|cite|improve this question

edited Mar 6 at 4:34

Robert Howard

2,3033935

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

share|cite|improve this question

share|cite|improve this question

edited Mar 6 at 4:34

Robert Howard

2,3033935

edited Mar 6 at 4:34

Robert Howard

2,3033935

edited Mar 6 at 4:34

Robert Howard

2,3033935

2,3033935

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

asked Feb 1 at 18:38

davidllerenavdavidllerenav

3228

3228

marked as duplicate by Clement C., Strants, Leucippus, YiFan, RRL Feb 2 at 6:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Clement C., Strants, Leucippus, YiFan, RRL Feb 2 at 6:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

add a comment | 

2 Answers
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• For (1), consider the function given by $f(0)=0$, and $f(x) = xsin frac{pi}{x}$ for $xneq 0$.




• For (2), consider the function defined by $f(0)=0$, and $f(x) = frac{x}{ln x}$ for $xneq 0$. (And recall that $sum_{n=2}^infty frac{1}{nln n} = infty$, e.g., by the integral test.)

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answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

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  $begingroup$
  Comments are not for extended discussion; this conversation has been moved to chat.
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  – Aloizio Macedo♦
  Feb 4 at 3:58
  

  
  
  
  

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The answer to (1) is clearly no. Let $f(x); x < 0$ be $-3x/2$ and $x>0$ be $f(x)=x^2$. Then $sum_n fleft(frac{1}{n}right) = sum_n frac{1}{n^2}$ converges, is continuous, but $f$ is not differentiable at $x=0$.

Another example [if the above example feels like 'cheating'] would be $f(x) = x^2left(sin left(frac{1}{x^3}right)+2right)$.

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edited Feb 1 at 19:56

answered Feb 1 at 19:40

MikeMike

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

• For (1), consider the function given by $f(0)=0$, and $f(x) = xsin frac{pi}{x}$ for $xneq 0$.




• For (2), consider the function defined by $f(0)=0$, and $f(x) = frac{x}{ln x}$ for $xneq 0$. (And recall that $sum_{n=2}^infty frac{1}{nln n} = infty$, e.g., by the integral test.)

share|cite|improve this answer

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comments are not for extended discussion; this conversation has been moved to chat.
  $endgroup$
  – Aloizio Macedo♦
  Feb 4 at 3:58
  

  
  
  
  

add a comment | 

0

$begingroup$

• For (1), consider the function given by $f(0)=0$, and $f(x) = xsin frac{pi}{x}$ for $xneq 0$.




• For (2), consider the function defined by $f(0)=0$, and $f(x) = frac{x}{ln x}$ for $xneq 0$. (And recall that $sum_{n=2}^infty frac{1}{nln n} = infty$, e.g., by the integral test.)

share|cite|improve this answer

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comments are not for extended discussion; this conversation has been moved to chat.
  $endgroup$
  – Aloizio Macedo♦
  Feb 4 at 3:58
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

• For (1), consider the function given by $f(0)=0$, and $f(x) = xsin frac{pi}{x}$ for $xneq 0$.




• For (2), consider the function defined by $f(0)=0$, and $f(x) = frac{x}{ln x}$ for $xneq 0$. (And recall that $sum_{n=2}^infty frac{1}{nln n} = infty$, e.g., by the integral test.)

share|cite|improve this answer

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

$endgroup$

• For (1), consider the function given by $f(0)=0$, and $f(x) = xsin frac{pi}{x}$ for $xneq 0$.




• For (2), consider the function defined by $f(0)=0$, and $f(x) = frac{x}{ln x}$ for $xneq 0$. (And recall that $sum_{n=2}^infty frac{1}{nln n} = infty$, e.g., by the integral test.)

share|cite|improve this answer

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

share|cite|improve this answer

share|cite|improve this answer

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

answered Feb 1 at 19:01

Clement C.Clement C.

51.1k34093

51.1k34093

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comments are not for extended discussion; this conversation has been moved to chat.
  $endgroup$
  – Aloizio Macedo♦
  Feb 4 at 3:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comments are not for extended discussion; this conversation has been moved to chat.
  $endgroup$
  – Aloizio Macedo♦
  Feb 4 at 3:58
  

  
  
  
  

$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Feb 4 at 3:58

$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Feb 4 at 3:58

add a comment | 

0

$begingroup$

The answer to (1) is clearly no. Let $f(x); x < 0$ be $-3x/2$ and $x>0$ be $f(x)=x^2$. Then $sum_n fleft(frac{1}{n}right) = sum_n frac{1}{n^2}$ converges, is continuous, but $f$ is not differentiable at $x=0$.

Another example [if the above example feels like 'cheating'] would be $f(x) = x^2left(sin left(frac{1}{x^3}right)+2right)$.

share|cite|improve this answer

edited Feb 1 at 19:56

answered Feb 1 at 19:40

MikeMike

4,611512

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add a comment | 

0

$begingroup$

The answer to (1) is clearly no. Let $f(x); x < 0$ be $-3x/2$ and $x>0$ be $f(x)=x^2$. Then $sum_n fleft(frac{1}{n}right) = sum_n frac{1}{n^2}$ converges, is continuous, but $f$ is not differentiable at $x=0$.

Another example [if the above example feels like 'cheating'] would be $f(x) = x^2left(sin left(frac{1}{x^3}right)+2right)$.

share|cite|improve this answer

edited Feb 1 at 19:56

answered Feb 1 at 19:40

MikeMike

4,611512

$endgroup$

add a comment | 

0

0

0

$begingroup$

The answer to (1) is clearly no. Let $f(x); x < 0$ be $-3x/2$ and $x>0$ be $f(x)=x^2$. Then $sum_n fleft(frac{1}{n}right) = sum_n frac{1}{n^2}$ converges, is continuous, but $f$ is not differentiable at $x=0$.

Another example [if the above example feels like 'cheating'] would be $f(x) = x^2left(sin left(frac{1}{x^3}right)+2right)$.

share|cite|improve this answer

edited Feb 1 at 19:56

answered Feb 1 at 19:40

MikeMike

4,611512

$endgroup$

The answer to (1) is clearly no. Let $f(x); x < 0$ be $-3x/2$ and $x>0$ be $f(x)=x^2$. Then $sum_n fleft(frac{1}{n}right) = sum_n frac{1}{n^2}$ converges, is continuous, but $f$ is not differentiable at $x=0$.

Another example [if the above example feels like 'cheating'] would be $f(x) = x^2left(sin left(frac{1}{x^3}right)+2right)$.

share|cite|improve this answer

edited Feb 1 at 19:56

answered Feb 1 at 19:40

MikeMike

4,611512

share|cite|improve this answer

share|cite|improve this answer

edited Feb 1 at 19:56

edited Feb 1 at 19:56

edited Feb 1 at 19:56

answered Feb 1 at 19:40

MikeMike

4,611512

answered Feb 1 at 19:40

MikeMike

4,611512

answered Feb 1 at 19:40

MikeMike

4,611512

4,611512

add a comment | 

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