A question on the limit $lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$

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I stumbled upon this question from a Calculus exam: Let $f in C^1(mathbbR)$ be monotonically decreasing such that $lim limits_x rightarrow infty f(x) = 0$. Prove that the limit
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$$
exists and is finite.

Now obviously the solution the writers of this problem had intended was to look at the integral
$$int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx = int limits_1^infty f(x)cos x , dx$$
which converges by the Dirichlet criterion and then the Riemann sums attributed to the partitions $Pi_n = 0, frac1n, frac2n, ..., 1
$ are $S_n=n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$ which if would imply
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2=int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx < infty$$
if the function $fraccos(frac1x)f(frac1x)x^2$ was Riemann-integrable in $[0,1]$.

The problem with that approach however, is that the integral
$$int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$
is not a Riemann-integral but an improper-integral so one cannot conclude that
$$lim limits_n rightarrow infty S_n = int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$

After many hours of thinking, trying to resolve this issue I realized that the question may be false as it makes no sense that the limit exists because that would sort of mean that the integral is a proper Riemann-integral which would imply the function is bounded (which is not necessarily true depending on the choice of $f$). To test this, I used Wolfram Alpha to calculate numerical estimates with the function $f(x)=frac1x$, which confirmed my speculations, though I have not been able to rigorously prove it.

My question is can you find an example that disproves the claim the question makes? Or was I wrong and the claim can be proven?

calculus

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edited Jul 31 at 12:36

asked Jul 30 at 23:11

Yuval Gamzon

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help...
  – Clement C.
  Jul 30 at 23:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But I have, the Dirichlet convergence test requires f to be monotonically decreasing
  – Yuval Gamzon
  Jul 31 at 4:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point, I had missed that.
  – Clement C.
  Jul 31 at 4:40
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

2

I stumbled upon this question from a Calculus exam: Let $f in C^1(mathbbR)$ be monotonically decreasing such that $lim limits_x rightarrow infty f(x) = 0$. Prove that the limit
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$$
exists and is finite.

Now obviously the solution the writers of this problem had intended was to look at the integral
$$int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx = int limits_1^infty f(x)cos x , dx$$
which converges by the Dirichlet criterion and then the Riemann sums attributed to the partitions $Pi_n = 0, frac1n, frac2n, ..., 1
$ are $S_n=n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$ which if would imply
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2=int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx < infty$$
if the function $fraccos(frac1x)f(frac1x)x^2$ was Riemann-integrable in $[0,1]$.

The problem with that approach however, is that the integral
$$int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$
is not a Riemann-integral but an improper-integral so one cannot conclude that
$$lim limits_n rightarrow infty S_n = int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$

After many hours of thinking, trying to resolve this issue I realized that the question may be false as it makes no sense that the limit exists because that would sort of mean that the integral is a proper Riemann-integral which would imply the function is bounded (which is not necessarily true depending on the choice of $f$). To test this, I used Wolfram Alpha to calculate numerical estimates with the function $f(x)=frac1x$, which confirmed my speculations, though I have not been able to rigorously prove it.

My question is can you find an example that disproves the claim the question makes? Or was I wrong and the claim can be proven?

calculus

share|cite|improve this question

edited Jul 31 at 12:36

asked Jul 30 at 23:11

Yuval Gamzon

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help...
  – Clement C.
  Jul 30 at 23:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But I have, the Dirichlet convergence test requires f to be monotonically decreasing
  – Yuval Gamzon
  Jul 31 at 4:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point, I had missed that.
  – Clement C.
  Jul 31 at 4:40
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

2

up vote
6
down vote

favorite

2

2

I stumbled upon this question from a Calculus exam: Let $f in C^1(mathbbR)$ be monotonically decreasing such that $lim limits_x rightarrow infty f(x) = 0$. Prove that the limit
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$$
exists and is finite.

Now obviously the solution the writers of this problem had intended was to look at the integral
$$int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx = int limits_1^infty f(x)cos x , dx$$
which converges by the Dirichlet criterion and then the Riemann sums attributed to the partitions $Pi_n = 0, frac1n, frac2n, ..., 1
$ are $S_n=n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$ which if would imply
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2=int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx < infty$$
if the function $fraccos(frac1x)f(frac1x)x^2$ was Riemann-integrable in $[0,1]$.

The problem with that approach however, is that the integral
$$int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$
is not a Riemann-integral but an improper-integral so one cannot conclude that
$$lim limits_n rightarrow infty S_n = int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$

After many hours of thinking, trying to resolve this issue I realized that the question may be false as it makes no sense that the limit exists because that would sort of mean that the integral is a proper Riemann-integral which would imply the function is bounded (which is not necessarily true depending on the choice of $f$). To test this, I used Wolfram Alpha to calculate numerical estimates with the function $f(x)=frac1x$, which confirmed my speculations, though I have not been able to rigorously prove it.

My question is can you find an example that disproves the claim the question makes? Or was I wrong and the claim can be proven?

calculus

share|cite|improve this question

edited Jul 31 at 12:36

asked Jul 30 at 23:11

Yuval Gamzon

424

I stumbled upon this question from a Calculus exam: Let $f in C^1(mathbbR)$ be monotonically decreasing such that $lim limits_x rightarrow infty f(x) = 0$. Prove that the limit
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$$
exists and is finite.

Now obviously the solution the writers of this problem had intended was to look at the integral
$$int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx = int limits_1^infty f(x)cos x , dx$$
which converges by the Dirichlet criterion and then the Riemann sums attributed to the partitions $Pi_n = 0, frac1n, frac2n, ..., 1
$ are $S_n=n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2$ which if would imply
$$lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj)f(fracnj)j^2=int limits_0^1 fraccos(frac1x)f(frac1x)x^2dx < infty$$
if the function $fraccos(frac1x)f(frac1x)x^2$ was Riemann-integrable in $[0,1]$.

The problem with that approach however, is that the integral
$$int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$
is not a Riemann-integral but an improper-integral so one cannot conclude that
$$lim limits_n rightarrow infty S_n = int limits_0^1fraccos(frac1x)f(frac1x)x^2dx$$

After many hours of thinking, trying to resolve this issue I realized that the question may be false as it makes no sense that the limit exists because that would sort of mean that the integral is a proper Riemann-integral which would imply the function is bounded (which is not necessarily true depending on the choice of $f$). To test this, I used Wolfram Alpha to calculate numerical estimates with the function $f(x)=frac1x$, which confirmed my speculations, though I have not been able to rigorously prove it.

My question is can you find an example that disproves the claim the question makes? Or was I wrong and the claim can be proven?

calculus

share|cite|improve this question

edited Jul 31 at 12:36

asked Jul 30 at 23:11

Yuval Gamzon

424

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 12:36

edited Jul 31 at 12:36

edited Jul 31 at 12:36

asked Jul 30 at 23:11

Yuval Gamzon

424

asked Jul 30 at 23:11

Yuval Gamzon

424

asked Jul 30 at 23:11

Yuval Gamzon

424

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help...
  – Clement C.
  Jul 30 at 23:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But I have, the Dirichlet convergence test requires f to be monotonically decreasing
  – Yuval Gamzon
  Jul 31 at 4:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point, I had missed that.
  – Clement C.
  Jul 31 at 4:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help...
  – Clement C.
  Jul 30 at 23:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But I have, the Dirichlet convergence test requires f to be monotonically decreasing
  – Yuval Gamzon
  Jul 31 at 4:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point, I had missed that.
  – Clement C.
  Jul 31 at 4:40
  

  
  
  
  

The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help...
– Clement C.
Jul 30 at 23:15

The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help...
– Clement C.
Jul 30 at 23:15

1

1

But I have, the Dirichlet convergence test requires f to be monotonically decreasing
– Yuval Gamzon
Jul 31 at 4:37

But I have, the Dirichlet convergence test requires f to be monotonically decreasing
– Yuval Gamzon
Jul 31 at 4:37

Good point, I had missed that.
– Clement C.
Jul 31 at 4:40

Good point, I had missed that.
– Clement C.
Jul 31 at 4:40

add a comment | 

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