Is there a function for which the sum converges while the limit diverges?
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Once I take my integral of f'(x) from n+1 to n I'll get f(n+1)-f(n). Since I already needed a function that's limit diverged, shouldn't my sum also diverge? This leads me to believe that this is a trick question and is impossible.
calculus sequences-and-series smooth-functions
edited Nov 22 at 2:45
Vee Hua Zhi
776124
asked Nov 22 at 1:20
Baq
111
add a comment |
up vote
2
down vote
favorite
Once I take my integral of f'(x) from n+1 to n I'll get f(n+1)-f(n). Since I already needed a function that's limit diverged, shouldn't my sum also diverge? This leads me to believe that this is a trick question and is impossible.
calculus sequences-and-series smooth-functions
edited Nov 22 at 2:45
Vee Hua Zhi
776124
asked Nov 22 at 1:20
Baq
111
1
I immediately want to go for a periodic sine function with 'wavelength' 1, such that the integral of the derivative is always 0 but it diverges. :) $sin(2pi x)$
– Milan Leonard
Nov 22 at 1:26
If I were to solve and set up my sum for f(x)=2pix then I would still get a divergent sum using divergence test.
– Baq
Nov 22 at 1:52
As you said, you're integral would be f(n+1) - f(n) but for all integers $sin(2pi x)$ is 0, so the sum from 1 to infinity of zero is still zero, but sine goes to zero.
– Milan Leonard
Nov 22 at 1:57
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Once I take my integral of f'(x) from n+1 to n I'll get f(n+1)-f(n). Since I already needed a function that's limit diverged, shouldn't my sum also diverge? This leads me to believe that this is a trick question and is impossible.
calculus sequences-and-series smooth-functions
edited Nov 22 at 2:45
Vee Hua Zhi
776124
asked Nov 22 at 1:20
Baq
111
Once I take my integral of f'(x) from n+1 to n I'll get f(n+1)-f(n). Since I already needed a function that's limit diverged, shouldn't my sum also diverge? This leads me to believe that this is a trick question and is impossible.
calculus sequences-and-series smooth-functions
calculus sequences-and-series smooth-functions
edited Nov 22 at 2:45
Vee Hua Zhi
776124
asked Nov 22 at 1:20
Baq
111
edited Nov 22 at 2:45
Vee Hua Zhi
776124
asked Nov 22 at 1:20
Baq
111
edited Nov 22 at 2:45
Vee Hua Zhi
776124
edited Nov 22 at 2:45
Vee Hua Zhi
776124
edited Nov 22 at 2:45
Vee Hua Zhi
776124
776124
asked Nov 22 at 1:20
Baq
111
asked Nov 22 at 1:20
Baq
111
asked Nov 22 at 1:20
Baq
111
111
1
I immediately want to go for a periodic sine function with 'wavelength' 1, such that the integral of the derivative is always 0 but it diverges. :) $sin(2pi x)$
– Milan Leonard
Nov 22 at 1:26
If I were to solve and set up my sum for f(x)=2pix then I would still get a divergent sum using divergence test.
– Baq
Nov 22 at 1:52
As you said, you're integral would be f(n+1) - f(n) but for all integers $sin(2pi x)$ is 0, so the sum from 1 to infinity of zero is still zero, but sine goes to zero.
– Milan Leonard
Nov 22 at 1:57
add a comment |
1
I immediately want to go for a periodic sine function with 'wavelength' 1, such that the integral of the derivative is always 0 but it diverges. :) $sin(2pi x)$
– Milan Leonard
Nov 22 at 1:26
If I were to solve and set up my sum for f(x)=2pix then I would still get a divergent sum using divergence test.
– Baq
Nov 22 at 1:52
As you said, you're integral would be f(n+1) - f(n) but for all integers $sin(2pi x)$ is 0, so the sum from 1 to infinity of zero is still zero, but sine goes to zero.
– Milan Leonard
Nov 22 at 1:57
1
1
I immediately want to go for a periodic sine function with 'wavelength' 1, such that the integral of the derivative is always 0 but it diverges. :) $sin(2pi x)$
– Milan Leonard
Nov 22 at 1:26
I immediately want to go for a periodic sine function with 'wavelength' 1, such that the integral of the derivative is always 0 but it diverges. :) $sin(2pi x)$
– Milan Leonard
Nov 22 at 1:26
If I were to solve and set up my sum for f(x)=2pix then I would still get a divergent sum using divergence test.
– Baq
Nov 22 at 1:52
If I were to solve and set up my sum for f(x)=2pix then I would still get a divergent sum using divergence test.
– Baq
Nov 22 at 1:52
As you said, you're integral would be f(n+1) - f(n) but for all integers $sin(2pi x)$ is 0, so the sum from 1 to infinity of zero is still zero, but sine goes to zero.
– Milan Leonard
Nov 22 at 1:57
As you said, you're integral would be f(n+1) - f(n) but for all integers $sin(2pi x)$ is 0, so the sum from 1 to infinity of zero is still zero, but sine goes to zero.
– Milan Leonard
Nov 22 at 1:57
add a comment |
1 Answer
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up vote
1
down vote
The upper expression simplifies to $lim_{n to infty} f(n)$.
So take $f(x) = sin(pi x)$.
$f(n) equiv 0$ but $f(x)$ oscillates, so it is divergent.
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
1
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
add a comment |
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1 Answer
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1 Answer
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up vote
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down vote
The upper expression simplifies to $lim_{n to infty} f(n)$.
So take $f(x) = sin(pi x)$.
$f(n) equiv 0$ but $f(x)$ oscillates, so it is divergent.
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
1
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
add a comment |
up vote
1
down vote
The upper expression simplifies to $lim_{n to infty} f(n)$.
So take $f(x) = sin(pi x)$.
$f(n) equiv 0$ but $f(x)$ oscillates, so it is divergent.
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
1
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
add a comment |
up vote
1
down vote
up vote
1
down vote
The upper expression simplifies to $lim_{n to infty} f(n)$.
So take $f(x) = sin(pi x)$.
$f(n) equiv 0$ but $f(x)$ oscillates, so it is divergent.
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
The upper expression simplifies to $lim_{n to infty} f(n)$.
So take $f(x) = sin(pi x)$.
$f(n) equiv 0$ but $f(x)$ oscillates, so it is divergent.
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
answered Nov 22 at 1:29
Lukas Kofler
1,2552519
1,2552519
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
1
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
add a comment |
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
1
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
how does the limit of f(n)= 0? Isn't sin(infinity) divergent?
– Baq
Nov 22 at 1:42
1
1
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
$lim_{ntoinfty} 0=0$, and the function evaluates to $0$ at every integer. This solution is a function that is zero at every integer, but does not have a limit at infinity.
– vadim123
Nov 22 at 2:49
add a comment |
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1
I immediately want to go for a periodic sine function with 'wavelength' 1, such that the integral of the derivative is always 0 but it diverges. :) $sin(2pi x)$
– Milan Leonard
Nov 22 at 1:26
If I were to solve and set up my sum for f(x)=2pix then I would still get a divergent sum using divergence test.
– Baq
Nov 22 at 1:52
As you said, you're integral would be f(n+1) - f(n) but for all integers $sin(2pi x)$ is 0, so the sum from 1 to infinity of zero is still zero, but sine goes to zero.
– Milan Leonard
Nov 22 at 1:57