Why is the inequality $sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}$ true?
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$
It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:
So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.
Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?
If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.
My claims come specifically from page 60 of this webpage from Dartmouth
integration sequences-and-series inequality summation power-series
add a comment |
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$
It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:
So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.
Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?
If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.
My claims come specifically from page 60 of this webpage from Dartmouth
integration sequences-and-series inequality summation power-series
Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47
add a comment |
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$
It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:
So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.
Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?
If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.
My claims come specifically from page 60 of this webpage from Dartmouth
integration sequences-and-series inequality summation power-series
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$
It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:
So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.
Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?
If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.
My claims come specifically from page 60 of this webpage from Dartmouth
integration sequences-and-series inequality summation power-series
integration sequences-and-series inequality summation power-series
asked Nov 27 '18 at 18:44
James Mitchell
26827
26827
Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47
add a comment |
Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47
Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47
Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47
add a comment |
4 Answers
4
active
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The right endpoint sums for the integral have the form:
$$
sum_{n = 2}^infty frac{1}{n^2}
$$
and we know:
$$
sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
frac{1}{n^2}$$
Multiplying by $-1$, we have:
$$
1 geq sum_{n = 1}^infty
frac{1}{n^2} -int_1^infty frac{1}{x^2}
$$
as we wanted.
add a comment |
A way to see that from the graph is as follows
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$
add a comment |
Hint: Subract $1$ from both sides to see inequality is the same as
$$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$
Now do your rectangle comparisons.
add a comment |
More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
The right endpoint sums for the integral have the form:
$$
sum_{n = 2}^infty frac{1}{n^2}
$$
and we know:
$$
sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
frac{1}{n^2}$$
Multiplying by $-1$, we have:
$$
1 geq sum_{n = 1}^infty
frac{1}{n^2} -int_1^infty frac{1}{x^2}
$$
as we wanted.
add a comment |
The right endpoint sums for the integral have the form:
$$
sum_{n = 2}^infty frac{1}{n^2}
$$
and we know:
$$
sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
frac{1}{n^2}$$
Multiplying by $-1$, we have:
$$
1 geq sum_{n = 1}^infty
frac{1}{n^2} -int_1^infty frac{1}{x^2}
$$
as we wanted.
add a comment |
The right endpoint sums for the integral have the form:
$$
sum_{n = 2}^infty frac{1}{n^2}
$$
and we know:
$$
sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
frac{1}{n^2}$$
Multiplying by $-1$, we have:
$$
1 geq sum_{n = 1}^infty
frac{1}{n^2} -int_1^infty frac{1}{x^2}
$$
as we wanted.
The right endpoint sums for the integral have the form:
$$
sum_{n = 2}^infty frac{1}{n^2}
$$
and we know:
$$
sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
frac{1}{n^2}$$
Multiplying by $-1$, we have:
$$
1 geq sum_{n = 1}^infty
frac{1}{n^2} -int_1^infty frac{1}{x^2}
$$
as we wanted.
answered Nov 27 '18 at 19:01
rubikscube09
1,169717
1,169717
add a comment |
add a comment |
A way to see that from the graph is as follows
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$
add a comment |
A way to see that from the graph is as follows
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$
add a comment |
A way to see that from the graph is as follows
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$
A way to see that from the graph is as follows
$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$
answered Nov 27 '18 at 18:47
gimusi
1
1
add a comment |
add a comment |
Hint: Subract $1$ from both sides to see inequality is the same as
$$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$
Now do your rectangle comparisons.
add a comment |
Hint: Subract $1$ from both sides to see inequality is the same as
$$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$
Now do your rectangle comparisons.
add a comment |
Hint: Subract $1$ from both sides to see inequality is the same as
$$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$
Now do your rectangle comparisons.
Hint: Subract $1$ from both sides to see inequality is the same as
$$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$
Now do your rectangle comparisons.
answered Nov 27 '18 at 19:09
zhw.
71.6k43075
71.6k43075
add a comment |
add a comment |
More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.
add a comment |
More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.
add a comment |
More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.
More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.
answered Nov 27 '18 at 19:12
J.G.
22.6k22136
22.6k22136
add a comment |
add a comment |
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Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47