Convergence of a series is wrong, using integral test

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Given the series : $$sum_1^inftye^-sqrtn$$ Determine if convergent or divergent.



The function is positive and monotonically decreasing function so I've used the "Integral Test"



$$int_1^inftye^-sqrtx$$



then:$$ e^-sqrtxleq e^sqrtxleq e^x$$



$$int_1^inftye^xdx$$ which is clearly Divergent but the answer for some reason is convergent



Edit: Sorry for the misinterpretation. I forgot that if f(x)>g(x) and f(x) is divergent, it doesn't necessarily mean that g(x) is divergent(contrary to convergent). Does anyone have a general direction for how do I suppose to solve it? I think that integral test is the most natural direction for solving this






calculus integration sequences-and-series convergence divergent-series







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  • You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral.
    – Michael Burr
    49 mins ago














up vote
1
down vote

favorite
1












Given the series : $$sum_1^inftye^-sqrtn$$ Determine if convergent or divergent.



The function is positive and monotonically decreasing function so I've used the "Integral Test"



$$int_1^inftye^-sqrtx$$



then:$$ e^-sqrtxleq e^sqrtxleq e^x$$



$$int_1^inftye^xdx$$ which is clearly Divergent but the answer for some reason is convergent



Edit: Sorry for the misinterpretation. I forgot that if f(x)>g(x) and f(x) is divergent, it doesn't necessarily mean that g(x) is divergent(contrary to convergent). Does anyone have a general direction for how do I suppose to solve it? I think that integral test is the most natural direction for solving this






calculus integration sequences-and-series convergence divergent-series







share|cite|improve this question























  • You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral.
    – Michael Burr
    49 mins ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Given the series : $$sum_1^inftye^-sqrtn$$ Determine if convergent or divergent.



The function is positive and monotonically decreasing function so I've used the "Integral Test"



$$int_1^inftye^-sqrtx$$



then:$$ e^-sqrtxleq e^sqrtxleq e^x$$



$$int_1^inftye^xdx$$ which is clearly Divergent but the answer for some reason is convergent



Edit: Sorry for the misinterpretation. I forgot that if f(x)>g(x) and f(x) is divergent, it doesn't necessarily mean that g(x) is divergent(contrary to convergent). Does anyone have a general direction for how do I suppose to solve it? I think that integral test is the most natural direction for solving this






calculus integration sequences-and-series convergence divergent-series







share|cite|improve this question















Given the series : $$sum_1^inftye^-sqrtn$$ Determine if convergent or divergent.



The function is positive and monotonically decreasing function so I've used the "Integral Test"



$$int_1^inftye^-sqrtx$$



then:$$ e^-sqrtxleq e^sqrtxleq e^x$$



$$int_1^inftye^xdx$$ which is clearly Divergent but the answer for some reason is convergent



Edit: Sorry for the misinterpretation. I forgot that if f(x)>g(x) and f(x) is divergent, it doesn't necessarily mean that g(x) is divergent(contrary to convergent). Does anyone have a general direction for how do I suppose to solve it? I think that integral test is the most natural direction for solving this






calculus integration sequences-and-series convergence divergent-series




calculus integration sequences-and-series convergence divergent-series






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edited 35 mins ago









José Carlos Santos

127k17102189




127k17102189










asked 55 mins ago









ProgC

562




562











  • You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral.
    – Michael Burr
    49 mins ago
















  • You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral.
    – Michael Burr
    49 mins ago















You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral.
– Michael Burr
49 mins ago




You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral.
– Michael Burr
49 mins ago










4 Answers
4






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up vote
3
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For each $ninmathbb N$, $e^-sqrt nleqslantsqrt ne^-sqrt n$. But the series $displaystylesum_n=1^inftysqrt ne^-sqrt n$ converges, by the integral test:$$int_1^inftysqrt xe^-sqrt x=lim_Mtoinftyleft[-left(2x+4sqrt x+4right)e^-sqrt xright]_1^M=frac 6e.$$Therefore, your series converges, too.






share|cite|improve this answer




















  • Thanks for the explanation
    – ProgC
    30 mins ago

















up vote
3
down vote













$u=sqrt x; du=frac12sqrt xdxRightarrow int_1^inftye^-sqrtxdx=2int_1^inftyue^-udu $ and this integral converges.






share|cite|improve this answer



























    up vote
    1
    down vote













    Hint:



    Near $+infty$, $;n^2=obigl(mathrm e^sqrt nbigr)$, whence $;mathrm e^-sqrt n=oBigl(dfrac1n^2Bigr)$.






    share|cite|improve this answer




















    • what those little zeros mean?
      – Jimmy Sabater
      39 mins ago










    • It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
      – Bernard
      35 mins ago











    • Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
      – Jimmy Sabater
      33 mins ago










    • Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
      – Bernard
      29 mins ago

















    up vote
    1
    down vote













    Hint.



    $$
    frac1e^sqrt nle frac1e^1.2ln n = frac1n^1.2
    $$






    share|cite|improve this answer




















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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      up vote
      3
      down vote













      For each $ninmathbb N$, $e^-sqrt nleqslantsqrt ne^-sqrt n$. But the series $displaystylesum_n=1^inftysqrt ne^-sqrt n$ converges, by the integral test:$$int_1^inftysqrt xe^-sqrt x=lim_Mtoinftyleft[-left(2x+4sqrt x+4right)e^-sqrt xright]_1^M=frac 6e.$$Therefore, your series converges, too.






      share|cite|improve this answer




















      • Thanks for the explanation
        – ProgC
        30 mins ago














      up vote
      3
      down vote













      For each $ninmathbb N$, $e^-sqrt nleqslantsqrt ne^-sqrt n$. But the series $displaystylesum_n=1^inftysqrt ne^-sqrt n$ converges, by the integral test:$$int_1^inftysqrt xe^-sqrt x=lim_Mtoinftyleft[-left(2x+4sqrt x+4right)e^-sqrt xright]_1^M=frac 6e.$$Therefore, your series converges, too.






      share|cite|improve this answer




















      • Thanks for the explanation
        – ProgC
        30 mins ago












      up vote
      3
      down vote










      up vote
      3
      down vote









      For each $ninmathbb N$, $e^-sqrt nleqslantsqrt ne^-sqrt n$. But the series $displaystylesum_n=1^inftysqrt ne^-sqrt n$ converges, by the integral test:$$int_1^inftysqrt xe^-sqrt x=lim_Mtoinftyleft[-left(2x+4sqrt x+4right)e^-sqrt xright]_1^M=frac 6e.$$Therefore, your series converges, too.






      share|cite|improve this answer












      For each $ninmathbb N$, $e^-sqrt nleqslantsqrt ne^-sqrt n$. But the series $displaystylesum_n=1^inftysqrt ne^-sqrt n$ converges, by the integral test:$$int_1^inftysqrt xe^-sqrt x=lim_Mtoinftyleft[-left(2x+4sqrt x+4right)e^-sqrt xright]_1^M=frac 6e.$$Therefore, your series converges, too.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 37 mins ago









      José Carlos Santos

      127k17102189




      127k17102189











      • Thanks for the explanation
        – ProgC
        30 mins ago
















      • Thanks for the explanation
        – ProgC
        30 mins ago















      Thanks for the explanation
      – ProgC
      30 mins ago




      Thanks for the explanation
      – ProgC
      30 mins ago










      up vote
      3
      down vote













      $u=sqrt x; du=frac12sqrt xdxRightarrow int_1^inftye^-sqrtxdx=2int_1^inftyue^-udu $ and this integral converges.






      share|cite|improve this answer
























        up vote
        3
        down vote













        $u=sqrt x; du=frac12sqrt xdxRightarrow int_1^inftye^-sqrtxdx=2int_1^inftyue^-udu $ and this integral converges.






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          $u=sqrt x; du=frac12sqrt xdxRightarrow int_1^inftye^-sqrtxdx=2int_1^inftyue^-udu $ and this integral converges.






          share|cite|improve this answer












          $u=sqrt x; du=frac12sqrt xdxRightarrow int_1^inftye^-sqrtxdx=2int_1^inftyue^-udu $ and this integral converges.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 37 mins ago









          Matematleta

          8,0692917




          8,0692917




















              up vote
              1
              down vote













              Hint:



              Near $+infty$, $;n^2=obigl(mathrm e^sqrt nbigr)$, whence $;mathrm e^-sqrt n=oBigl(dfrac1n^2Bigr)$.






              share|cite|improve this answer




















              • what those little zeros mean?
                – Jimmy Sabater
                39 mins ago










              • It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
                – Bernard
                35 mins ago











              • Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
                – Jimmy Sabater
                33 mins ago










              • Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
                – Bernard
                29 mins ago














              up vote
              1
              down vote













              Hint:



              Near $+infty$, $;n^2=obigl(mathrm e^sqrt nbigr)$, whence $;mathrm e^-sqrt n=oBigl(dfrac1n^2Bigr)$.






              share|cite|improve this answer




















              • what those little zeros mean?
                – Jimmy Sabater
                39 mins ago










              • It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
                – Bernard
                35 mins ago











              • Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
                – Jimmy Sabater
                33 mins ago










              • Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
                – Bernard
                29 mins ago












              up vote
              1
              down vote










              up vote
              1
              down vote









              Hint:



              Near $+infty$, $;n^2=obigl(mathrm e^sqrt nbigr)$, whence $;mathrm e^-sqrt n=oBigl(dfrac1n^2Bigr)$.






              share|cite|improve this answer












              Hint:



              Near $+infty$, $;n^2=obigl(mathrm e^sqrt nbigr)$, whence $;mathrm e^-sqrt n=oBigl(dfrac1n^2Bigr)$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 43 mins ago









              Bernard

              113k636104




              113k636104











              • what those little zeros mean?
                – Jimmy Sabater
                39 mins ago










              • It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
                – Bernard
                35 mins ago











              • Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
                – Jimmy Sabater
                33 mins ago










              • Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
                – Bernard
                29 mins ago
















              • what those little zeros mean?
                – Jimmy Sabater
                39 mins ago










              • It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
                – Bernard
                35 mins ago











              • Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
                – Jimmy Sabater
                33 mins ago










              • Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
                – Bernard
                29 mins ago















              what those little zeros mean?
              – Jimmy Sabater
              39 mins ago




              what those little zeros mean?
              – Jimmy Sabater
              39 mins ago












              It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
              – Bernard
              35 mins ago





              It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $dfrac fg$ tends to $0$ at $infty$.
              – Bernard
              35 mins ago













              Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
              – Jimmy Sabater
              33 mins ago




              Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer
              – Jimmy Sabater
              33 mins ago












              Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
              – Bernard
              29 mins ago




              Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school.
              – Bernard
              29 mins ago










              up vote
              1
              down vote













              Hint.



              $$
              frac1e^sqrt nle frac1e^1.2ln n = frac1n^1.2
              $$






              share|cite|improve this answer
























                up vote
                1
                down vote













                Hint.



                $$
                frac1e^sqrt nle frac1e^1.2ln n = frac1n^1.2
                $$






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Hint.



                  $$
                  frac1e^sqrt nle frac1e^1.2ln n = frac1n^1.2
                  $$






                  share|cite|improve this answer












                  Hint.



                  $$
                  frac1e^sqrt nle frac1e^1.2ln n = frac1n^1.2
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 12 mins ago









                  Cesareo

                  6,6172414




                  6,6172414



























                       

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