The value of $a$ besides $1$ for which $gamma...
$begingroup$
I know that Euler-Mascheroni constant is given by
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k}-int_{1}^{n}frac{1}{x}dxright)$$
I am not sure whether there exists a value $a$ besides $1$ such that the equation below holds.
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k^a}-int_{1}^{n}frac{1}{x^a}dxright)$$
I did some calculation for $0<a<1$, but as I set $n$ bigger and bigger, all the results seem to fall apart. For example, when $n=9999$, the result seems fine.
However, as $n$ gets larger, Wolfram Alpha starts to give me two different results. I am not sure whether it's because there is just no such value as $a$ (I am confident there is one), or is Wolfram Alpha doing the wrong calculation. I need some assistance.
calculus sequences-and-series
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
$endgroup$
|
show 6 more comments
$begingroup$
I know that Euler-Mascheroni constant is given by
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k}-int_{1}^{n}frac{1}{x}dxright)$$
I am not sure whether there exists a value $a$ besides $1$ such that the equation below holds.
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k^a}-int_{1}^{n}frac{1}{x^a}dxright)$$
I did some calculation for $0<a<1$, but as I set $n$ bigger and bigger, all the results seem to fall apart. For example, when $n=9999$, the result seems fine.
However, as $n$ gets larger, Wolfram Alpha starts to give me two different results. I am not sure whether it's because there is just no such value as $a$ (I am confident there is one), or is Wolfram Alpha doing the wrong calculation. I need some assistance.
calculus sequences-and-series
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
$endgroup$
1
$begingroup$
The limit equals $frac{1}{1-a}+zeta(a)$.
$endgroup$
– Kemono Chen
Dec 11 '18 at 1:33
$begingroup$
Good idea! I see what you are saying. Let me try that first. If you are right, I will try to find a proof for your claim.
$endgroup$
– Larry
Dec 11 '18 at 2:38
$begingroup$
This question may be useful.
$endgroup$
– Kemono Chen
Dec 11 '18 at 7:24
$begingroup$
So I tried your proposal of $frac{1}{1-a}+zeta{(a)}$. It seems like only 1 satisfy the limit. There are no other values, but I am still not convinced
$endgroup$
– Larry
Dec 12 '18 at 23:01
1
$begingroup$
Catalan $approx 0.915966$ is a constant. I used it to force Mathematica assume a "variable" is in $(0,1)$.
$endgroup$
– Kemono Chen
Dec 14 '18 at 0:24
|
show 6 more comments
$begingroup$
I know that Euler-Mascheroni constant is given by
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k}-int_{1}^{n}frac{1}{x}dxright)$$
I am not sure whether there exists a value $a$ besides $1$ such that the equation below holds.
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k^a}-int_{1}^{n}frac{1}{x^a}dxright)$$
I did some calculation for $0<a<1$, but as I set $n$ bigger and bigger, all the results seem to fall apart. For example, when $n=9999$, the result seems fine.
However, as $n$ gets larger, Wolfram Alpha starts to give me two different results. I am not sure whether it's because there is just no such value as $a$ (I am confident there is one), or is Wolfram Alpha doing the wrong calculation. I need some assistance.
calculus sequences-and-series
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
$endgroup$
I know that Euler-Mascheroni constant is given by
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k}-int_{1}^{n}frac{1}{x}dxright)$$
I am not sure whether there exists a value $a$ besides $1$ such that the equation below holds.
$$gamma =lim_{nrightarrowinfty}left(sum_{k=1}^{n}frac{1}{k^a}-int_{1}^{n}frac{1}{x^a}dxright)$$
I did some calculation for $0<a<1$, but as I set $n$ bigger and bigger, all the results seem to fall apart. For example, when $n=9999$, the result seems fine.
However, as $n$ gets larger, Wolfram Alpha starts to give me two different results. I am not sure whether it's because there is just no such value as $a$ (I am confident there is one), or is Wolfram Alpha doing the wrong calculation. I need some assistance.
calculus sequences-and-series
calculus sequences-and-series
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
asked Dec 11 '18 at 0:11
LarryLarry
2,39131129
2,39131129
1
$begingroup$
The limit equals $frac{1}{1-a}+zeta(a)$.
$endgroup$
– Kemono Chen
Dec 11 '18 at 1:33
$begingroup$
Good idea! I see what you are saying. Let me try that first. If you are right, I will try to find a proof for your claim.
$endgroup$
– Larry
Dec 11 '18 at 2:38
$begingroup$
This question may be useful.
$endgroup$
– Kemono Chen
Dec 11 '18 at 7:24
$begingroup$
So I tried your proposal of $frac{1}{1-a}+zeta{(a)}$. It seems like only 1 satisfy the limit. There are no other values, but I am still not convinced
$endgroup$
– Larry
Dec 12 '18 at 23:01
1
$begingroup$
Catalan $approx 0.915966$ is a constant. I used it to force Mathematica assume a "variable" is in $(0,1)$.
$endgroup$
– Kemono Chen
Dec 14 '18 at 0:24
|
show 6 more comments
1
$begingroup$
The limit equals $frac{1}{1-a}+zeta(a)$.
$endgroup$
– Kemono Chen
Dec 11 '18 at 1:33
$begingroup$
Good idea! I see what you are saying. Let me try that first. If you are right, I will try to find a proof for your claim.
$endgroup$
– Larry
Dec 11 '18 at 2:38
$begingroup$
This question may be useful.
$endgroup$
– Kemono Chen
Dec 11 '18 at 7:24
$begingroup$
So I tried your proposal of $frac{1}{1-a}+zeta{(a)}$. It seems like only 1 satisfy the limit. There are no other values, but I am still not convinced
$endgroup$
– Larry
Dec 12 '18 at 23:01
1
$begingroup$
Catalan $approx 0.915966$ is a constant. I used it to force Mathematica assume a "variable" is in $(0,1)$.
$endgroup$
– Kemono Chen
Dec 14 '18 at 0:24
1
1
$begingroup$
The limit equals $frac{1}{1-a}+zeta(a)$.
$endgroup$
– Kemono Chen
Dec 11 '18 at 1:33
$begingroup$
The limit equals $frac{1}{1-a}+zeta(a)$.
$endgroup$
– Kemono Chen
Dec 11 '18 at 1:33
$begingroup$
Good idea! I see what you are saying. Let me try that first. If you are right, I will try to find a proof for your claim.
$endgroup$
– Larry
Dec 11 '18 at 2:38
$begingroup$
Good idea! I see what you are saying. Let me try that first. If you are right, I will try to find a proof for your claim.
$endgroup$
– Larry
Dec 11 '18 at 2:38
$begingroup$
This question may be useful.
$endgroup$
– Kemono Chen
Dec 11 '18 at 7:24
$begingroup$
This question may be useful.
$endgroup$
– Kemono Chen
Dec 11 '18 at 7:24
$begingroup$
So I tried your proposal of $frac{1}{1-a}+zeta{(a)}$. It seems like only 1 satisfy the limit. There are no other values, but I am still not convinced
$endgroup$
– Larry
Dec 12 '18 at 23:01
$begingroup$
So I tried your proposal of $frac{1}{1-a}+zeta{(a)}$. It seems like only 1 satisfy the limit. There are no other values, but I am still not convinced
$endgroup$
– Larry
Dec 12 '18 at 23:01
1
1
$begingroup$
Catalan $approx 0.915966$ is a constant. I used it to force Mathematica assume a "variable" is in $(0,1)$.
$endgroup$
– Kemono Chen
Dec 14 '18 at 0:24
$begingroup$
Catalan $approx 0.915966$ is a constant. I used it to force Mathematica assume a "variable" is in $(0,1)$.
$endgroup$
– Kemono Chen
Dec 14 '18 at 0:24
|
show 6 more comments
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1
$begingroup$
The limit equals $frac{1}{1-a}+zeta(a)$.
$endgroup$
– Kemono Chen
Dec 11 '18 at 1:33
$begingroup$
Good idea! I see what you are saying. Let me try that first. If you are right, I will try to find a proof for your claim.
$endgroup$
– Larry
Dec 11 '18 at 2:38
$begingroup$
This question may be useful.
$endgroup$
– Kemono Chen
Dec 11 '18 at 7:24
$begingroup$
So I tried your proposal of $frac{1}{1-a}+zeta{(a)}$. It seems like only 1 satisfy the limit. There are no other values, but I am still not convinced
$endgroup$
– Larry
Dec 12 '18 at 23:01
1
$begingroup$
Catalan $approx 0.915966$ is a constant. I used it to force Mathematica assume a "variable" is in $(0,1)$.
$endgroup$
– Kemono Chen
Dec 14 '18 at 0:24