$sum_n=1^infty frac1n^z$ diverges if $0 <Re(z) < 1$
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I have been trying to solve this problem for more that two days and yet i have no idea of what to do. I am trying to show that $sum_n=1^infty frac1n^z$ diverges if 0 < $Re(z) < 1$.
If $z$ is real it turns out to be very simple because
$int_n^n+1 fracdtt^z leq frac1n^z$ and $int_1^infty fracdtt^z$ diverges.
But for arbitrary $z in mathbbC$ the same idea turns out in a mess, the hint is: analyze $sum_n^infty big(frac1n^z - int_n^n+1 fracdtt^z big) $
complex-analysis
asked Aug 11 at 0:54
Alicia Basilio
746
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up vote
3
down vote
favorite
I have been trying to solve this problem for more that two days and yet i have no idea of what to do. I am trying to show that $sum_n=1^infty frac1n^z$ diverges if 0 < $Re(z) < 1$.
If $z$ is real it turns out to be very simple because
$int_n^n+1 fracdtt^z leq frac1n^z$ and $int_1^infty fracdtt^z$ diverges.
But for arbitrary $z in mathbbC$ the same idea turns out in a mess, the hint is: analyze $sum_n^infty big(frac1n^z - int_n^n+1 fracdtt^z big) $
complex-analysis
asked Aug 11 at 0:54
Alicia Basilio
746
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have been trying to solve this problem for more that two days and yet i have no idea of what to do. I am trying to show that $sum_n=1^infty frac1n^z$ diverges if 0 < $Re(z) < 1$.
If $z$ is real it turns out to be very simple because
$int_n^n+1 fracdtt^z leq frac1n^z$ and $int_1^infty fracdtt^z$ diverges.
But for arbitrary $z in mathbbC$ the same idea turns out in a mess, the hint is: analyze $sum_n^infty big(frac1n^z - int_n^n+1 fracdtt^z big) $
complex-analysis
asked Aug 11 at 0:54
Alicia Basilio
746
I have been trying to solve this problem for more that two days and yet i have no idea of what to do. I am trying to show that $sum_n=1^infty frac1n^z$ diverges if 0 < $Re(z) < 1$.
If $z$ is real it turns out to be very simple because
$int_n^n+1 fracdtt^z leq frac1n^z$ and $int_1^infty fracdtt^z$ diverges.
But for arbitrary $z in mathbbC$ the same idea turns out in a mess, the hint is: analyze $sum_n^infty big(frac1n^z - int_n^n+1 fracdtt^z big) $
complex-analysis
asked Aug 11 at 0:54
Alicia Basilio
746
edited Aug 11 at 1:55
asked Aug 11 at 0:54
Alicia Basilio
746
asked Aug 11 at 0:54
Alicia Basilio
746
asked Aug 11 at 0:54
Alicia Basilio
746
746
add a comment |Â
add a comment |Â
1 Answer
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up vote
2
down vote
accepted
I think your hint should be "analyse
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)".$$
Integrating by parts,
$$frac1n^z-int_n^n+1fracdtt^z
=zint_n^n+1fracn+1-tt^z+1,dt.$$
If $z=x+iy$ then this integral is $O(n^-x-1)$. As $x>0$.
the series
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)$$
converges absolutely, but
$$sum_n=1^inftyint_n^n+1fracdtt^z$$
diverges, and so
$$sum_n=1^inftyfrac1n^z$$
must diverge.
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I think your hint should be "analyse
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)".$$
Integrating by parts,
$$frac1n^z-int_n^n+1fracdtt^z
=zint_n^n+1fracn+1-tt^z+1,dt.$$
If $z=x+iy$ then this integral is $O(n^-x-1)$. As $x>0$.
the series
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)$$
converges absolutely, but
$$sum_n=1^inftyint_n^n+1fracdtt^z$$
diverges, and so
$$sum_n=1^inftyfrac1n^z$$
must diverge.
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
add a comment |Â
up vote
2
down vote
accepted
I think your hint should be "analyse
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)".$$
Integrating by parts,
$$frac1n^z-int_n^n+1fracdtt^z
=zint_n^n+1fracn+1-tt^z+1,dt.$$
If $z=x+iy$ then this integral is $O(n^-x-1)$. As $x>0$.
the series
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)$$
converges absolutely, but
$$sum_n=1^inftyint_n^n+1fracdtt^z$$
diverges, and so
$$sum_n=1^inftyfrac1n^z$$
must diverge.
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think your hint should be "analyse
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)".$$
Integrating by parts,
$$frac1n^z-int_n^n+1fracdtt^z
=zint_n^n+1fracn+1-tt^z+1,dt.$$
If $z=x+iy$ then this integral is $O(n^-x-1)$. As $x>0$.
the series
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)$$
converges absolutely, but
$$sum_n=1^inftyint_n^n+1fracdtt^z$$
diverges, and so
$$sum_n=1^inftyfrac1n^z$$
must diverge.
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
I think your hint should be "analyse
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)".$$
Integrating by parts,
$$frac1n^z-int_n^n+1fracdtt^z
=zint_n^n+1fracn+1-tt^z+1,dt.$$
If $z=x+iy$ then this integral is $O(n^-x-1)$. As $x>0$.
the series
$$sum_n=1^inftyleft(frac1n^z-int_n^n+1fracdtt^zright)$$
converges absolutely, but
$$sum_n=1^inftyint_n^n+1fracdtt^z$$
diverges, and so
$$sum_n=1^inftyfrac1n^z$$
must diverge.
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
answered Aug 11 at 1:46
Lord Shark the Unknown
86.5k952112
86.5k952112
add a comment |Â
add a comment |Â
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