Mixing
Series interpretation of integral
$begingroup$
I'm currently stuck with the following question:
Prove, that $ln(2) = limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k}$ by rewriting the left side as an Integral.
So my current thoughts are:
$ln(2) = int_1^2 frac{1}{x} mathrm{d}x$
$limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k} = limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1} = limlimits_{n to 0} sum_{k=1}^{frac{2-1}{n}} n frac{1}{nk + 1}$
Like this, the sum is rewritten as a step function with the width of each equidistant step decreasing. If the sum is the upper sum, I have to show, that $frac{1}{nk+1}=ln(1 + frac{k+1}{n})$. How can I "get rid" if the $ln$?
real-analysis riemann-sum
asked Feb 1 at 17:16
TimTim
246
$endgroup$
add a comment |
$begingroup$
I'm currently stuck with the following question:
Prove, that $ln(2) = limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k}$ by rewriting the left side as an Integral.
So my current thoughts are:
$ln(2) = int_1^2 frac{1}{x} mathrm{d}x$
$limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k} = limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1} = limlimits_{n to 0} sum_{k=1}^{frac{2-1}{n}} n frac{1}{nk + 1}$
Like this, the sum is rewritten as a step function with the width of each equidistant step decreasing. If the sum is the upper sum, I have to show, that $frac{1}{nk+1}=ln(1 + frac{k+1}{n})$. How can I "get rid" if the $ln$?
real-analysis riemann-sum
asked Feb 1 at 17:16
TimTim
246
$endgroup$
$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Feb 1 at 17:51
add a comment |
$begingroup$
I'm currently stuck with the following question:
Prove, that $ln(2) = limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k}$ by rewriting the left side as an Integral.
So my current thoughts are:
$ln(2) = int_1^2 frac{1}{x} mathrm{d}x$
$limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k} = limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1} = limlimits_{n to 0} sum_{k=1}^{frac{2-1}{n}} n frac{1}{nk + 1}$
Like this, the sum is rewritten as a step function with the width of each equidistant step decreasing. If the sum is the upper sum, I have to show, that $frac{1}{nk+1}=ln(1 + frac{k+1}{n})$. How can I "get rid" if the $ln$?
real-analysis riemann-sum
asked Feb 1 at 17:16
TimTim
246
$endgroup$
I'm currently stuck with the following question:
Prove, that $ln(2) = limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k}$ by rewriting the left side as an Integral.
So my current thoughts are:
$ln(2) = int_1^2 frac{1}{x} mathrm{d}x$
$limlimits_{n to infty} sum_{k=1}^n frac{1}{n+k} = limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1} = limlimits_{n to 0} sum_{k=1}^{frac{2-1}{n}} n frac{1}{nk + 1}$
Like this, the sum is rewritten as a step function with the width of each equidistant step decreasing. If the sum is the upper sum, I have to show, that $frac{1}{nk+1}=ln(1 + frac{k+1}{n})$. How can I "get rid" if the $ln$?
real-analysis riemann-sum
real-analysis riemann-sum
asked Feb 1 at 17:16
TimTim
246
asked Feb 1 at 17:16
TimTim
246
asked Feb 1 at 17:16
TimTim
246
asked Feb 1 at 17:16
TimTim
246
asked Feb 1 at 17:16
TimTim
246
246
$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Feb 1 at 17:51
add a comment |
$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Feb 1 at 17:51
$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Feb 1 at 17:51
$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Feb 1 at 17:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,ldots, 2$, then
$$limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1}= limlimits_{n to infty} sum_{k=1}^n left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right] frac{1}{1+frac{k}{n}}=int_1^2frac{1}{x}dx.$$
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
$endgroup$
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
|
show 9 more comments
$begingroup$
hint
Write it as
$$frac{b-a}{n}sum_{k=1}^nfleft(a+kfrac{b-a}{n}right)$$
the limit will be
$$int_a^bf(x)dx$$
In your case, $a=0; ; b=1, f(x)=frac{1}{1+x}$
Or
$a=1,;b=2 ;$ and$ ; f(x)=frac 1x.$
edited Feb 1 at 17:34
J.G.
33.2k23252
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,ldots, 2$, then
$$limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1}= limlimits_{n to infty} sum_{k=1}^n left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right] frac{1}{1+frac{k}{n}}=int_1^2frac{1}{x}dx.$$
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
$endgroup$
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
|
show 9 more comments
$begingroup$
You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,ldots, 2$, then
$$limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1}= limlimits_{n to infty} sum_{k=1}^n left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right] frac{1}{1+frac{k}{n}}=int_1^2frac{1}{x}dx.$$
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
$endgroup$
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
|
show 9 more comments
$begingroup$
You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,ldots, 2$, then
$$limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1}= limlimits_{n to infty} sum_{k=1}^n left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right] frac{1}{1+frac{k}{n}}=int_1^2frac{1}{x}dx.$$
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
$endgroup$
You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,ldots, 2$, then
$$limlimits_{n to infty} sum_{k=1}^n frac{1}{n} frac{1}{frac{k}{n} + 1}= limlimits_{n to infty} sum_{k=1}^n left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right] frac{1}{1+frac{k}{n}}=int_1^2frac{1}{x}dx.$$
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
answered Feb 1 at 17:28
Mars PlasticMars Plastic
1,455122
1,455122
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
|
show 9 more comments
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
So $left[left(1+frac{k}{n}right)-left(1+frac{k-1}{n}right)right]$ is the "width" of the subdivision and $frac{1}{1+ frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral?
$endgroup$
– Tim
Feb 1 at 17:43
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums.
$endgroup$
– Mars Plastic
Feb 1 at 17:48
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition?
$endgroup$
– Tim
Feb 1 at 18:15
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
OK, so how is the integral introduced (defined) in your lecture?
$endgroup$
– Mars Plastic
Feb 2 at 1:20
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
$begingroup$
We have introduced the integral with step functions: Let $varphi$ be a step function, then the integral of $varphi$ over $[a,b]$ is defined as $int_a^b varphi(x) dx = sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(varphi_n)$ is uniformly converging to $f$, then $int_a^b f(x) dx = limlimits{n to infty}int_a^b varphi_n(x) dx$.
$endgroup$
– Tim
Feb 2 at 6:53
|
show 9 more comments
$begingroup$
hint
Write it as
$$frac{b-a}{n}sum_{k=1}^nfleft(a+kfrac{b-a}{n}right)$$
the limit will be
$$int_a^bf(x)dx$$
In your case, $a=0; ; b=1, f(x)=frac{1}{1+x}$
Or
$a=1,;b=2 ;$ and$ ; f(x)=frac 1x.$
edited Feb 1 at 17:34
J.G.
33.2k23252
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
add a comment |
$begingroup$
hint
Write it as
$$frac{b-a}{n}sum_{k=1}^nfleft(a+kfrac{b-a}{n}right)$$
the limit will be
$$int_a^bf(x)dx$$
In your case, $a=0; ; b=1, f(x)=frac{1}{1+x}$
Or
$a=1,;b=2 ;$ and$ ; f(x)=frac 1x.$
edited Feb 1 at 17:34
J.G.
33.2k23252
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
add a comment |
$begingroup$
hint
Write it as
$$frac{b-a}{n}sum_{k=1}^nfleft(a+kfrac{b-a}{n}right)$$
the limit will be
$$int_a^bf(x)dx$$
In your case, $a=0; ; b=1, f(x)=frac{1}{1+x}$
Or
$a=1,;b=2 ;$ and$ ; f(x)=frac 1x.$
edited Feb 1 at 17:34
J.G.
33.2k23252
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
hint
Write it as
$$frac{b-a}{n}sum_{k=1}^nfleft(a+kfrac{b-a}{n}right)$$
the limit will be
$$int_a^bf(x)dx$$
In your case, $a=0; ; b=1, f(x)=frac{1}{1+x}$
Or
$a=1,;b=2 ;$ and$ ; f(x)=frac 1x.$
edited Feb 1 at 17:34
J.G.
33.2k23252
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
edited Feb 1 at 17:34
J.G.
33.2k23252
edited Feb 1 at 17:34
J.G.
33.2k23252
edited Feb 1 at 17:34
J.G.
33.2k23252
33.2k23252
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Feb 1 at 17:21
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
add a comment |
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
$begingroup$
So it would be $int_1^2 frac{1}{x} dx = frac{1}{n} sum_{k=1}^n frac{1}{1 + frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me?
$endgroup$
– Tim
Feb 1 at 17:29
add a comment |
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$begingroup$
See math.stackexchange.com/questions/469885/…
$endgroup$
– lab bhattacharjee
Feb 1 at 17:51