Using series for determining the convergence of improper integral
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We apply the integral test for determining the convergence of series . For example integral test is useful for $sum_{n=1}^infty frac{1}{n}$ or $sum_{n=1}^infty frac{ln n}{n}$ . Because the integral test is biconditional , I thought about the other way . I mean determining the convergence of the improper integral with the help of series . Is it practical ? What are the examples for it ?
real-analysis calculus integration sequences-and-series improper-integrals
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add a comment |
$begingroup$
We apply the integral test for determining the convergence of series . For example integral test is useful for $sum_{n=1}^infty frac{1}{n}$ or $sum_{n=1}^infty frac{ln n}{n}$ . Because the integral test is biconditional , I thought about the other way . I mean determining the convergence of the improper integral with the help of series . Is it practical ? What are the examples for it ?
real-analysis calculus integration sequences-and-series improper-integrals
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$$int_1^inftyfrac{dx}{x^2}$$
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– DonAntonio
Dec 23 '18 at 22:13
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@DonAntonio Thanks but we can easily solve that integral without help of series . I'm interested in cases in which solving integral by itself is not possible and then we use series .
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– S.H.W
Dec 23 '18 at 22:15
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it can be used sometimes, however it is not so practical. The reasons is that there are more tools to use to calculate integrals than tools to calculate series. Moreover: the comparison just works in the case of non-negative integrands, but there are convergent improper integrals that are not absolutely convergent
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– Masacroso
Dec 23 '18 at 22:18
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@Masacroso Yes , I agree with you . It was just interesting to think about that .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
$begingroup$
We apply the integral test for determining the convergence of series . For example integral test is useful for $sum_{n=1}^infty frac{1}{n}$ or $sum_{n=1}^infty frac{ln n}{n}$ . Because the integral test is biconditional , I thought about the other way . I mean determining the convergence of the improper integral with the help of series . Is it practical ? What are the examples for it ?
real-analysis calculus integration sequences-and-series improper-integrals
$endgroup$
We apply the integral test for determining the convergence of series . For example integral test is useful for $sum_{n=1}^infty frac{1}{n}$ or $sum_{n=1}^infty frac{ln n}{n}$ . Because the integral test is biconditional , I thought about the other way . I mean determining the convergence of the improper integral with the help of series . Is it practical ? What are the examples for it ?
real-analysis calculus integration sequences-and-series improper-integrals
real-analysis calculus integration sequences-and-series improper-integrals
edited Dec 23 '18 at 22:09
S.H.W
asked Dec 23 '18 at 22:02
S.H.WS.H.W
1,2181923
1,2181923
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$$int_1^inftyfrac{dx}{x^2}$$
$endgroup$
– DonAntonio
Dec 23 '18 at 22:13
$begingroup$
@DonAntonio Thanks but we can easily solve that integral without help of series . I'm interested in cases in which solving integral by itself is not possible and then we use series .
$endgroup$
– S.H.W
Dec 23 '18 at 22:15
$begingroup$
it can be used sometimes, however it is not so practical. The reasons is that there are more tools to use to calculate integrals than tools to calculate series. Moreover: the comparison just works in the case of non-negative integrands, but there are convergent improper integrals that are not absolutely convergent
$endgroup$
– Masacroso
Dec 23 '18 at 22:18
$begingroup$
@Masacroso Yes , I agree with you . It was just interesting to think about that .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
$begingroup$
$$int_1^inftyfrac{dx}{x^2}$$
$endgroup$
– DonAntonio
Dec 23 '18 at 22:13
$begingroup$
@DonAntonio Thanks but we can easily solve that integral without help of series . I'm interested in cases in which solving integral by itself is not possible and then we use series .
$endgroup$
– S.H.W
Dec 23 '18 at 22:15
$begingroup$
it can be used sometimes, however it is not so practical. The reasons is that there are more tools to use to calculate integrals than tools to calculate series. Moreover: the comparison just works in the case of non-negative integrands, but there are convergent improper integrals that are not absolutely convergent
$endgroup$
– Masacroso
Dec 23 '18 at 22:18
$begingroup$
@Masacroso Yes , I agree with you . It was just interesting to think about that .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
$begingroup$
$$int_1^inftyfrac{dx}{x^2}$$
$endgroup$
– DonAntonio
Dec 23 '18 at 22:13
$begingroup$
$$int_1^inftyfrac{dx}{x^2}$$
$endgroup$
– DonAntonio
Dec 23 '18 at 22:13
$begingroup$
@DonAntonio Thanks but we can easily solve that integral without help of series . I'm interested in cases in which solving integral by itself is not possible and then we use series .
$endgroup$
– S.H.W
Dec 23 '18 at 22:15
$begingroup$
@DonAntonio Thanks but we can easily solve that integral without help of series . I'm interested in cases in which solving integral by itself is not possible and then we use series .
$endgroup$
– S.H.W
Dec 23 '18 at 22:15
$begingroup$
it can be used sometimes, however it is not so practical. The reasons is that there are more tools to use to calculate integrals than tools to calculate series. Moreover: the comparison just works in the case of non-negative integrands, but there are convergent improper integrals that are not absolutely convergent
$endgroup$
– Masacroso
Dec 23 '18 at 22:18
$begingroup$
it can be used sometimes, however it is not so practical. The reasons is that there are more tools to use to calculate integrals than tools to calculate series. Moreover: the comparison just works in the case of non-negative integrands, but there are convergent improper integrals that are not absolutely convergent
$endgroup$
– Masacroso
Dec 23 '18 at 22:18
$begingroup$
@Masacroso Yes , I agree with you . It was just interesting to think about that .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
$begingroup$
@Masacroso Yes , I agree with you . It was just interesting to think about that .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
1 Answer
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Sure. In some cases, it is practical. Consider, for instance, the integral $displaystyleint_2^inftyfrac1{x(x-1)},mathrm dx$. Thenbegin{align}int_2^inftyfrac1{x(x-1)},mathrm dxtext{ converges}&iffsum_{n=2}^inftyfrac1{n(n-1)}text{ converges}\&iffsum_{n=2}^inftyleft(frac1n-frac1{n+1}right)text{ converges,}end{align}and it is clear that the last statment holds (it's a telescoping series).
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Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
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Right now, I can't think of an example.
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– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Sure. In some cases, it is practical. Consider, for instance, the integral $displaystyleint_2^inftyfrac1{x(x-1)},mathrm dx$. Thenbegin{align}int_2^inftyfrac1{x(x-1)},mathrm dxtext{ converges}&iffsum_{n=2}^inftyfrac1{n(n-1)}text{ converges}\&iffsum_{n=2}^inftyleft(frac1n-frac1{n+1}right)text{ converges,}end{align}and it is clear that the last statment holds (it's a telescoping series).
$endgroup$
$begingroup$
Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
$begingroup$
Right now, I can't think of an example.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
$begingroup$
Sure. In some cases, it is practical. Consider, for instance, the integral $displaystyleint_2^inftyfrac1{x(x-1)},mathrm dx$. Thenbegin{align}int_2^inftyfrac1{x(x-1)},mathrm dxtext{ converges}&iffsum_{n=2}^inftyfrac1{n(n-1)}text{ converges}\&iffsum_{n=2}^inftyleft(frac1n-frac1{n+1}right)text{ converges,}end{align}and it is clear that the last statment holds (it's a telescoping series).
$endgroup$
$begingroup$
Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
$begingroup$
Right now, I can't think of an example.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
$begingroup$
Sure. In some cases, it is practical. Consider, for instance, the integral $displaystyleint_2^inftyfrac1{x(x-1)},mathrm dx$. Thenbegin{align}int_2^inftyfrac1{x(x-1)},mathrm dxtext{ converges}&iffsum_{n=2}^inftyfrac1{n(n-1)}text{ converges}\&iffsum_{n=2}^inftyleft(frac1n-frac1{n+1}right)text{ converges,}end{align}and it is clear that the last statment holds (it's a telescoping series).
$endgroup$
Sure. In some cases, it is practical. Consider, for instance, the integral $displaystyleint_2^inftyfrac1{x(x-1)},mathrm dx$. Thenbegin{align}int_2^inftyfrac1{x(x-1)},mathrm dxtext{ converges}&iffsum_{n=2}^inftyfrac1{n(n-1)}text{ converges}\&iffsum_{n=2}^inftyleft(frac1n-frac1{n+1}right)text{ converges,}end{align}and it is clear that the last statment holds (it's a telescoping series).
answered Dec 23 '18 at 22:18
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
$begingroup$
Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
$begingroup$
Right now, I can't think of an example.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
$begingroup$
Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
$begingroup$
Right now, I can't think of an example.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
$begingroup$
Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
$begingroup$
Thanks , it was interesting example . Do you know a case in which solving integral by itself is not possible and then we use series for determining convergence of integral ?
$endgroup$
– S.H.W
Dec 23 '18 at 22:23
$begingroup$
Right now, I can't think of an example.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Right now, I can't think of an example.
$endgroup$
– José Carlos Santos
Dec 23 '18 at 22:25
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
$begingroup$
Okay , thanks again .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32
add a comment |
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$begingroup$
$$int_1^inftyfrac{dx}{x^2}$$
$endgroup$
– DonAntonio
Dec 23 '18 at 22:13
$begingroup$
@DonAntonio Thanks but we can easily solve that integral without help of series . I'm interested in cases in which solving integral by itself is not possible and then we use series .
$endgroup$
– S.H.W
Dec 23 '18 at 22:15
$begingroup$
it can be used sometimes, however it is not so practical. The reasons is that there are more tools to use to calculate integrals than tools to calculate series. Moreover: the comparison just works in the case of non-negative integrands, but there are convergent improper integrals that are not absolutely convergent
$endgroup$
– Masacroso
Dec 23 '18 at 22:18
$begingroup$
@Masacroso Yes , I agree with you . It was just interesting to think about that .
$endgroup$
– S.H.W
Dec 23 '18 at 22:32