What value does $frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots(2n)}$ tend to? [duplicate]












2















This question already has an answer here:




  • How to prove that $lim frac{1}{n} sqrt[n]{(n+1)(n+2)… 2n} = frac{4}{e}$

    5 answers




I need to find where this sequence tends to:
$$frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}$$



My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.






sequences-and-series







share|cite|improve this question















marked as duplicate by Arnaud D., Martin Sleziak, Don Thousand, amWhy, José Carlos Santos Nov 21 '18 at 23:52


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    Which part of the expression is under the root? And which part is in the denominator?
    – Yuriy S
    Nov 10 '18 at 15:00












  • all of it after the root times the 1/n
    – stelioball
    Nov 10 '18 at 15:00












  • Please use the following code: frac{numerator}{denominator} and sqrt[n]{expression}
    – Yuriy S
    Nov 10 '18 at 15:01










  • I’ve edited it. Use braces for roots, exponents, subscripts, etc. Also, using fractions is better.
    – KM101
    Nov 10 '18 at 15:01








  • 3




    As the $n$-th root is clearly $>n$, there's no way the sequence can tend to zero.
    – Lord Shark the Unknown
    Nov 10 '18 at 15:12
















2















This question already has an answer here:




  • How to prove that $lim frac{1}{n} sqrt[n]{(n+1)(n+2)… 2n} = frac{4}{e}$

    5 answers




I need to find where this sequence tends to:
$$frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}$$



My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.






sequences-and-series







share|cite|improve this question















marked as duplicate by Arnaud D., Martin Sleziak, Don Thousand, amWhy, José Carlos Santos Nov 21 '18 at 23:52


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    Which part of the expression is under the root? And which part is in the denominator?
    – Yuriy S
    Nov 10 '18 at 15:00












  • all of it after the root times the 1/n
    – stelioball
    Nov 10 '18 at 15:00












  • Please use the following code: frac{numerator}{denominator} and sqrt[n]{expression}
    – Yuriy S
    Nov 10 '18 at 15:01










  • I’ve edited it. Use braces for roots, exponents, subscripts, etc. Also, using fractions is better.
    – KM101
    Nov 10 '18 at 15:01








  • 3




    As the $n$-th root is clearly $>n$, there's no way the sequence can tend to zero.
    – Lord Shark the Unknown
    Nov 10 '18 at 15:12














2












2








2








This question already has an answer here:




  • How to prove that $lim frac{1}{n} sqrt[n]{(n+1)(n+2)… 2n} = frac{4}{e}$

    5 answers




I need to find where this sequence tends to:
$$frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}$$



My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.






sequences-and-series







share|cite|improve this question
















This question already has an answer here:




  • How to prove that $lim frac{1}{n} sqrt[n]{(n+1)(n+2)… 2n} = frac{4}{e}$

    5 answers




I need to find where this sequence tends to:
$$frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}$$



My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.





This question already has an answer here:




  • How to prove that $lim frac{1}{n} sqrt[n]{(n+1)(n+2)… 2n} = frac{4}{e}$

    5 answers







sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 11 '18 at 13:35







stelioball

















asked Nov 10 '18 at 14:56









stelioballstelioball

355




355




marked as duplicate by Arnaud D., Martin Sleziak, Don Thousand, amWhy, José Carlos Santos Nov 21 '18 at 23:52


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Arnaud D., Martin Sleziak, Don Thousand, amWhy, José Carlos Santos Nov 21 '18 at 23:52


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    Which part of the expression is under the root? And which part is in the denominator?
    – Yuriy S
    Nov 10 '18 at 15:00












  • all of it after the root times the 1/n
    – stelioball
    Nov 10 '18 at 15:00












  • Please use the following code: frac{numerator}{denominator} and sqrt[n]{expression}
    – Yuriy S
    Nov 10 '18 at 15:01










  • I’ve edited it. Use braces for roots, exponents, subscripts, etc. Also, using fractions is better.
    – KM101
    Nov 10 '18 at 15:01








  • 3




    As the $n$-th root is clearly $>n$, there's no way the sequence can tend to zero.
    – Lord Shark the Unknown
    Nov 10 '18 at 15:12














  • 1




    Which part of the expression is under the root? And which part is in the denominator?
    – Yuriy S
    Nov 10 '18 at 15:00












  • all of it after the root times the 1/n
    – stelioball
    Nov 10 '18 at 15:00












  • Please use the following code: frac{numerator}{denominator} and sqrt[n]{expression}
    – Yuriy S
    Nov 10 '18 at 15:01










  • I’ve edited it. Use braces for roots, exponents, subscripts, etc. Also, using fractions is better.
    – KM101
    Nov 10 '18 at 15:01








  • 3




    As the $n$-th root is clearly $>n$, there's no way the sequence can tend to zero.
    – Lord Shark the Unknown
    Nov 10 '18 at 15:12








1




1




Which part of the expression is under the root? And which part is in the denominator?
– Yuriy S
Nov 10 '18 at 15:00






Which part of the expression is under the root? And which part is in the denominator?
– Yuriy S
Nov 10 '18 at 15:00














all of it after the root times the 1/n
– stelioball
Nov 10 '18 at 15:00






all of it after the root times the 1/n
– stelioball
Nov 10 '18 at 15:00














Please use the following code: frac{numerator}{denominator} and sqrt[n]{expression}
– Yuriy S
Nov 10 '18 at 15:01




Please use the following code: frac{numerator}{denominator} and sqrt[n]{expression}
– Yuriy S
Nov 10 '18 at 15:01












I’ve edited it. Use braces for roots, exponents, subscripts, etc. Also, using fractions is better.
– KM101
Nov 10 '18 at 15:01






I’ve edited it. Use braces for roots, exponents, subscripts, etc. Also, using fractions is better.
– KM101
Nov 10 '18 at 15:01






3




3




As the $n$-th root is clearly $>n$, there's no way the sequence can tend to zero.
– Lord Shark the Unknown
Nov 10 '18 at 15:12




As the $n$-th root is clearly $>n$, there's no way the sequence can tend to zero.
– Lord Shark the Unknown
Nov 10 '18 at 15:12










3 Answers
3






active

oldest

votes


















4














By ratio root criterion we have



$$a_n=frac{n(n+1)(n+2)...(2n)}{n^n} quad b_n=frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}=sqrt[n] a_n$$



then



$$frac{a_{n+1}}{a_n}=frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}frac{n^n}{(n+1)^{n+1}}=frac{(2n+2)(2n+1)}{n(n+1)}frac{1}{left(1+frac1nright)^n}to frac 4 e$$



which implies that $b_n to frac 4 e$.






share|cite|improve this answer



















  • 1




    Gimusi.Very nice!
    – Peter Szilas
    Nov 10 '18 at 17:08










  • Thanks my dear friend! :)
    – gimusi
    Nov 10 '18 at 17:21



















4














$$ a_n
= frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots (2n)}
= sqrt[n]{1(1+frac{1}{n})(1+frac{2}{n})cdots (1+frac{n}{n})}.
$$
$$
log a_n
= frac{1}{n} { log 1 + log (1+frac{1}{n}) + log (1+frac{2}{n})
+ cdots + log (1+frac{n}{n}) }.
$$



We know the above equations. Hence,



$$
limlimits_{n rightarrow infty} log a_n = int_1^2 log x dx
= [xlog x -x ]_1^2 = 2log 2 - 1 = log frac{4}{e}.
$$



Thus, the limit of $a_n$ is $frac{4}{e}$.






share|cite|improve this answer





















  • Are you using Riemann sum?
    – gimusi
    Nov 10 '18 at 15:27










  • @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
    – Doyun Nam
    Nov 10 '18 at 15:33










  • Maybe you could explicitly refer to it to make it clear.
    – gimusi
    Nov 10 '18 at 16:17



















3














By Stirling's formula,
begin{align}
lnleft[(n+1)(n+2)cdots(2n)right]
&=ln (2n)!-ln n!=(2n+1/2)ln(2n)-(n+1/2)ln n-n+o(1)\
&=(2n+1/2)ln 2+nln n-n+o(1)
end{align}
and so
$$lnleft[n(n+1)(n+2)cdots(2n)right]
=(2n+1/2)ln 2+(n+1)ln n-n+o(1).$$
Then
$$lnsqrt[n]{n(n+1)(n+2)cdots(2n)}=2ln 2+ln n-1+o(1)$$
and
$$lnfrac{sqrt[n]{n(n+1)(n+2)cdots(2n)}}n=2ln 2-1+o(1).$$
We conclude that your sequence converges to $4/e$.






share|cite|improve this answer




























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    By ratio root criterion we have



    $$a_n=frac{n(n+1)(n+2)...(2n)}{n^n} quad b_n=frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}=sqrt[n] a_n$$



    then



    $$frac{a_{n+1}}{a_n}=frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}frac{n^n}{(n+1)^{n+1}}=frac{(2n+2)(2n+1)}{n(n+1)}frac{1}{left(1+frac1nright)^n}to frac 4 e$$



    which implies that $b_n to frac 4 e$.






    share|cite|improve this answer



















    • 1




      Gimusi.Very nice!
      – Peter Szilas
      Nov 10 '18 at 17:08










    • Thanks my dear friend! :)
      – gimusi
      Nov 10 '18 at 17:21
















    4














    By ratio root criterion we have



    $$a_n=frac{n(n+1)(n+2)...(2n)}{n^n} quad b_n=frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}=sqrt[n] a_n$$



    then



    $$frac{a_{n+1}}{a_n}=frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}frac{n^n}{(n+1)^{n+1}}=frac{(2n+2)(2n+1)}{n(n+1)}frac{1}{left(1+frac1nright)^n}to frac 4 e$$



    which implies that $b_n to frac 4 e$.






    share|cite|improve this answer



















    • 1




      Gimusi.Very nice!
      – Peter Szilas
      Nov 10 '18 at 17:08










    • Thanks my dear friend! :)
      – gimusi
      Nov 10 '18 at 17:21














    4












    4








    4






    By ratio root criterion we have



    $$a_n=frac{n(n+1)(n+2)...(2n)}{n^n} quad b_n=frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}=sqrt[n] a_n$$



    then



    $$frac{a_{n+1}}{a_n}=frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}frac{n^n}{(n+1)^{n+1}}=frac{(2n+2)(2n+1)}{n(n+1)}frac{1}{left(1+frac1nright)^n}to frac 4 e$$



    which implies that $b_n to frac 4 e$.






    share|cite|improve this answer














    By ratio root criterion we have



    $$a_n=frac{n(n+1)(n+2)...(2n)}{n^n} quad b_n=frac{1}{n} sqrt[n]{n(n+1)(n+2)...(2n)}=sqrt[n] a_n$$



    then



    $$frac{a_{n+1}}{a_n}=frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}frac{n^n}{(n+1)^{n+1}}=frac{(2n+2)(2n+1)}{n(n+1)}frac{1}{left(1+frac1nright)^n}to frac 4 e$$



    which implies that $b_n to frac 4 e$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 21 '18 at 18:01

























    answered Nov 10 '18 at 15:20









    gimusigimusi

    1




    1








    • 1




      Gimusi.Very nice!
      – Peter Szilas
      Nov 10 '18 at 17:08










    • Thanks my dear friend! :)
      – gimusi
      Nov 10 '18 at 17:21














    • 1




      Gimusi.Very nice!
      – Peter Szilas
      Nov 10 '18 at 17:08










    • Thanks my dear friend! :)
      – gimusi
      Nov 10 '18 at 17:21








    1




    1




    Gimusi.Very nice!
    – Peter Szilas
    Nov 10 '18 at 17:08




    Gimusi.Very nice!
    – Peter Szilas
    Nov 10 '18 at 17:08












    Thanks my dear friend! :)
    – gimusi
    Nov 10 '18 at 17:21




    Thanks my dear friend! :)
    – gimusi
    Nov 10 '18 at 17:21











    4














    $$ a_n
    = frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots (2n)}
    = sqrt[n]{1(1+frac{1}{n})(1+frac{2}{n})cdots (1+frac{n}{n})}.
    $$
    $$
    log a_n
    = frac{1}{n} { log 1 + log (1+frac{1}{n}) + log (1+frac{2}{n})
    + cdots + log (1+frac{n}{n}) }.
    $$



    We know the above equations. Hence,



    $$
    limlimits_{n rightarrow infty} log a_n = int_1^2 log x dx
    = [xlog x -x ]_1^2 = 2log 2 - 1 = log frac{4}{e}.
    $$



    Thus, the limit of $a_n$ is $frac{4}{e}$.






    share|cite|improve this answer





















    • Are you using Riemann sum?
      – gimusi
      Nov 10 '18 at 15:27










    • @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
      – Doyun Nam
      Nov 10 '18 at 15:33










    • Maybe you could explicitly refer to it to make it clear.
      – gimusi
      Nov 10 '18 at 16:17
















    4














    $$ a_n
    = frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots (2n)}
    = sqrt[n]{1(1+frac{1}{n})(1+frac{2}{n})cdots (1+frac{n}{n})}.
    $$
    $$
    log a_n
    = frac{1}{n} { log 1 + log (1+frac{1}{n}) + log (1+frac{2}{n})
    + cdots + log (1+frac{n}{n}) }.
    $$



    We know the above equations. Hence,



    $$
    limlimits_{n rightarrow infty} log a_n = int_1^2 log x dx
    = [xlog x -x ]_1^2 = 2log 2 - 1 = log frac{4}{e}.
    $$



    Thus, the limit of $a_n$ is $frac{4}{e}$.






    share|cite|improve this answer





















    • Are you using Riemann sum?
      – gimusi
      Nov 10 '18 at 15:27










    • @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
      – Doyun Nam
      Nov 10 '18 at 15:33










    • Maybe you could explicitly refer to it to make it clear.
      – gimusi
      Nov 10 '18 at 16:17














    4












    4








    4






    $$ a_n
    = frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots (2n)}
    = sqrt[n]{1(1+frac{1}{n})(1+frac{2}{n})cdots (1+frac{n}{n})}.
    $$
    $$
    log a_n
    = frac{1}{n} { log 1 + log (1+frac{1}{n}) + log (1+frac{2}{n})
    + cdots + log (1+frac{n}{n}) }.
    $$



    We know the above equations. Hence,



    $$
    limlimits_{n rightarrow infty} log a_n = int_1^2 log x dx
    = [xlog x -x ]_1^2 = 2log 2 - 1 = log frac{4}{e}.
    $$



    Thus, the limit of $a_n$ is $frac{4}{e}$.






    share|cite|improve this answer












    $$ a_n
    = frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots (2n)}
    = sqrt[n]{1(1+frac{1}{n})(1+frac{2}{n})cdots (1+frac{n}{n})}.
    $$
    $$
    log a_n
    = frac{1}{n} { log 1 + log (1+frac{1}{n}) + log (1+frac{2}{n})
    + cdots + log (1+frac{n}{n}) }.
    $$



    We know the above equations. Hence,



    $$
    limlimits_{n rightarrow infty} log a_n = int_1^2 log x dx
    = [xlog x -x ]_1^2 = 2log 2 - 1 = log frac{4}{e}.
    $$



    Thus, the limit of $a_n$ is $frac{4}{e}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 10 '18 at 15:17









    Doyun NamDoyun Nam

    41619




    41619












    • Are you using Riemann sum?
      – gimusi
      Nov 10 '18 at 15:27










    • @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
      – Doyun Nam
      Nov 10 '18 at 15:33










    • Maybe you could explicitly refer to it to make it clear.
      – gimusi
      Nov 10 '18 at 16:17


















    • Are you using Riemann sum?
      – gimusi
      Nov 10 '18 at 15:27










    • @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
      – Doyun Nam
      Nov 10 '18 at 15:33










    • Maybe you could explicitly refer to it to make it clear.
      – gimusi
      Nov 10 '18 at 16:17
















    Are you using Riemann sum?
    – gimusi
    Nov 10 '18 at 15:27




    Are you using Riemann sum?
    – gimusi
    Nov 10 '18 at 15:27












    @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
    – Doyun Nam
    Nov 10 '18 at 15:33




    @gimusi Yes, $log x$ is continuous, and so Riemann integrable.
    – Doyun Nam
    Nov 10 '18 at 15:33












    Maybe you could explicitly refer to it to make it clear.
    – gimusi
    Nov 10 '18 at 16:17




    Maybe you could explicitly refer to it to make it clear.
    – gimusi
    Nov 10 '18 at 16:17











    3














    By Stirling's formula,
    begin{align}
    lnleft[(n+1)(n+2)cdots(2n)right]
    &=ln (2n)!-ln n!=(2n+1/2)ln(2n)-(n+1/2)ln n-n+o(1)\
    &=(2n+1/2)ln 2+nln n-n+o(1)
    end{align}
    and so
    $$lnleft[n(n+1)(n+2)cdots(2n)right]
    =(2n+1/2)ln 2+(n+1)ln n-n+o(1).$$
    Then
    $$lnsqrt[n]{n(n+1)(n+2)cdots(2n)}=2ln 2+ln n-1+o(1)$$
    and
    $$lnfrac{sqrt[n]{n(n+1)(n+2)cdots(2n)}}n=2ln 2-1+o(1).$$
    We conclude that your sequence converges to $4/e$.






    share|cite|improve this answer


























      3














      By Stirling's formula,
      begin{align}
      lnleft[(n+1)(n+2)cdots(2n)right]
      &=ln (2n)!-ln n!=(2n+1/2)ln(2n)-(n+1/2)ln n-n+o(1)\
      &=(2n+1/2)ln 2+nln n-n+o(1)
      end{align}
      and so
      $$lnleft[n(n+1)(n+2)cdots(2n)right]
      =(2n+1/2)ln 2+(n+1)ln n-n+o(1).$$
      Then
      $$lnsqrt[n]{n(n+1)(n+2)cdots(2n)}=2ln 2+ln n-1+o(1)$$
      and
      $$lnfrac{sqrt[n]{n(n+1)(n+2)cdots(2n)}}n=2ln 2-1+o(1).$$
      We conclude that your sequence converges to $4/e$.






      share|cite|improve this answer
























        3












        3








        3






        By Stirling's formula,
        begin{align}
        lnleft[(n+1)(n+2)cdots(2n)right]
        &=ln (2n)!-ln n!=(2n+1/2)ln(2n)-(n+1/2)ln n-n+o(1)\
        &=(2n+1/2)ln 2+nln n-n+o(1)
        end{align}
        and so
        $$lnleft[n(n+1)(n+2)cdots(2n)right]
        =(2n+1/2)ln 2+(n+1)ln n-n+o(1).$$
        Then
        $$lnsqrt[n]{n(n+1)(n+2)cdots(2n)}=2ln 2+ln n-1+o(1)$$
        and
        $$lnfrac{sqrt[n]{n(n+1)(n+2)cdots(2n)}}n=2ln 2-1+o(1).$$
        We conclude that your sequence converges to $4/e$.






        share|cite|improve this answer












        By Stirling's formula,
        begin{align}
        lnleft[(n+1)(n+2)cdots(2n)right]
        &=ln (2n)!-ln n!=(2n+1/2)ln(2n)-(n+1/2)ln n-n+o(1)\
        &=(2n+1/2)ln 2+nln n-n+o(1)
        end{align}
        and so
        $$lnleft[n(n+1)(n+2)cdots(2n)right]
        =(2n+1/2)ln 2+(n+1)ln n-n+o(1).$$
        Then
        $$lnsqrt[n]{n(n+1)(n+2)cdots(2n)}=2ln 2+ln n-1+o(1)$$
        and
        $$lnfrac{sqrt[n]{n(n+1)(n+2)cdots(2n)}}n=2ln 2-1+o(1).$$
        We conclude that your sequence converges to $4/e$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 10 '18 at 15:17









        Lord Shark the UnknownLord Shark the Unknown

        102k959132




        102k959132















            Popular posts from this blog

            MongoDB - Not Authorized To Execute Command

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
            Read more

            Npm cannot find a required file even through it is in the searched directory

            Image
            up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
            Read more

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Image
            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
            Read more