Using series for determining the convergence of improper integral












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$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$int_1^inftyfrac{dx}{x^2}$$
    $endgroup$
    – 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$int_1^inftyfrac{dx}{x^2}$$
    $endgroup$
    – 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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












  • $begingroup$
    $$int_1^inftyfrac{dx}{x^2}$$
    $endgroup$
    – 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










  • $begingroup$
    @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










1 Answer
1






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oldest

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2












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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).






share|cite|improve this answer









$endgroup$



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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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