Methods to show the convergence/divergence of following series:












0












$begingroup$


This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?






real-analysis







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  • $begingroup$
    Use Cauchy-Condensation test...
    $endgroup$
    – RAM_3R
    Jan 17 at 13:50
















0












$begingroup$


This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Use Cauchy-Condensation test...
    $endgroup$
    – RAM_3R
    Jan 17 at 13:50














0












0








0


1



$begingroup$


This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?






real-analysis







share|cite|improve this question









$endgroup$




This is the series I have a question about: $sum_{n=1}^{infty} frac{(ln n)^2 }{2^{ln n}}$.
The root and ratio test are both inconclusive, so I tried Raabe's test of convergence. It seems too difficult to find the limit because the term $frac{x_n}{x_{n+1}}$ becomes huge (also when applying de l'Hopital's rule). Could someone help me please? Are there any rules/methods on how to approach proving the convergence of series? How can they be applied on this given series?






real-analysis




real-analysis






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asked Jan 17 at 13:17









ZacharyZachary

3229




3229












  • $begingroup$
    Use Cauchy-Condensation test...
    $endgroup$
    – RAM_3R
    Jan 17 at 13:50


















  • $begingroup$
    Use Cauchy-Condensation test...
    $endgroup$
    – RAM_3R
    Jan 17 at 13:50
















$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50




$begingroup$
Use Cauchy-Condensation test...
$endgroup$
– RAM_3R
Jan 17 at 13:50










3 Answers
3






active

oldest

votes


















0












$begingroup$

$a^{ln n} = e^{ln a ln n} =n^{ln a}$
so the sum
$sum (ln n)^b/a^{ln n}$
converges if
$ln a > 1$,
diverges if
$ln a < 1$,
and depends on the value of $b$
if $ln a=1$.



In this case,
$ln a =ln 2 < 1$
so the sum diverges.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$



    Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$



    Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.






      share|cite|improve this answer









      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        $a^{ln n} = e^{ln a ln n} =n^{ln a}$
        so the sum
        $sum (ln n)^b/a^{ln n}$
        converges if
        $ln a > 1$,
        diverges if
        $ln a < 1$,
        and depends on the value of $b$
        if $ln a=1$.



        In this case,
        $ln a =ln 2 < 1$
        so the sum diverges.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          $a^{ln n} = e^{ln a ln n} =n^{ln a}$
          so the sum
          $sum (ln n)^b/a^{ln n}$
          converges if
          $ln a > 1$,
          diverges if
          $ln a < 1$,
          and depends on the value of $b$
          if $ln a=1$.



          In this case,
          $ln a =ln 2 < 1$
          so the sum diverges.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            $a^{ln n} = e^{ln a ln n} =n^{ln a}$
            so the sum
            $sum (ln n)^b/a^{ln n}$
            converges if
            $ln a > 1$,
            diverges if
            $ln a < 1$,
            and depends on the value of $b$
            if $ln a=1$.



            In this case,
            $ln a =ln 2 < 1$
            so the sum diverges.






            share|cite|improve this answer









            $endgroup$



            $a^{ln n} = e^{ln a ln n} =n^{ln a}$
            so the sum
            $sum (ln n)^b/a^{ln n}$
            converges if
            $ln a > 1$,
            diverges if
            $ln a < 1$,
            and depends on the value of $b$
            if $ln a=1$.



            In this case,
            $ln a =ln 2 < 1$
            so the sum diverges.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 17 at 13:39









            marty cohenmarty cohen

            75.5k549130




            75.5k549130























                0












                $begingroup$

                Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$



                Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$



                Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$



                  Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$



                  Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$



                    Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$



                    Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges






                    share|cite|improve this answer











                    $endgroup$



                    Let $a_n=sum(ln n)^2/2^{ln n}=sum(ln n)^2/n^{ln 2}$



                    Now $s_n=sum frac{2^n .(ln 2^n)^2}{2^{nln 2}}=sum 2^{n(1-ln 2)}n^2(ln 2)^2$



                    Since $1-ln 2>0$, So $s_n$ diverges $Rightarrow$ $a_n$ diverges







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 17 at 16:17

























                    answered Jan 17 at 13:57









                    RAM_3RRAM_3R

                    602216




                    602216























                        0












                        $begingroup$

                        There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.






                            share|cite|improve this answer









                            $endgroup$



                            There is this useful Cauchy condensation test, implying that $sum_{n=1}^{infty}a_n$ is convergent iff $sum_{n=1}^{infty}2^na_{2^n}$ is convergent, usable if $a_n$ is nonincreasing sequence of nonnegative numbers. Since $2^{ln n}$ is faster than $(ln$ $n)^2$ one can assume that this sequence is nonincreasing ( as it is nonincreasing for $n$ large enough ), and of course $a_n$ are nonnegative, so one can use this test. It follows that our sum converges iff sum $sum_{n=1}^{infty}2^n(ln(2^n))^2/2^{ln(2^n)}$ = $sum_{n=1}^{infty}2^n(nln2)^2/2^{nln2}$ = $sum_{n=1}^{infty}n^2(ln2)^2/2^{n(ln2 - 1)}$ = $sum_{n=1}^{infty}n^2(ln2)^22^{n(1-ln2)}$ converges, but this one obviously diverges, as $ln2$ $< 1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 17 at 16:42









                            Jakub AndruszkiewiczJakub Andruszkiewicz

                            2116




                            2116






























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