Test the convergence of the series












-2














For $a>0$, test the convergence of the infinite series: $$sum a^{ln(n)}$$ Which test should be better to use here? I tried the D'Alembert's ratio test. $$lim_{nto infty} {frac{a^{ln(n)}}{a^{ln(n+1)}}}= lim_{nto infty}a^{ln{frac{n}{n+1}}}=a^0=1$$ So the test failed.



UPDATE: I observed that $${a^{ln(n)}}=e^{{ln(a)}{ln(n)}}={n^{ln(a)}}={{1over n^{ln({1over a})}} }$$Further by test of auxiliary series $$sum frac{1}{n^p}$$ we know that criterion for convergence is $p>1$.
Here $p=ln({1over a})$.Hence the series will be convergent when $ln(a)<-1$ i.e. when $0<a<1/e$, since $a>0$ is already given.Finally this solved my question.






sequences-and-series







share|cite|improve this question




















  • 1




    Which ones did you try already? Which ones do you know? What were the results? We aren't here to do your homework, you know.
    – 5xum
    Nov 14 '17 at 9:47












  • That's essentially the same as this question
    – Professor Vector
    Nov 14 '17 at 9:52






  • 1




    this converges if $$log(|a|)+1<0$$ is hold.
    – Dr. Sonnhard Graubner
    Nov 14 '17 at 9:58










  • @5xum I tried the D Alembert ratio test but it failed.
    – Shatabdi Sinha
    Nov 14 '17 at 10:07






  • 1




    Make yourself familiar with Raabe's test and the other ones on that page. But here, note that $a^{ln n} = expbigl((ln n)(ln a)bigr) = n^{ln a}$.
    – Daniel Fischer♦
    Nov 14 '17 at 10:26


















-2














For $a>0$, test the convergence of the infinite series: $$sum a^{ln(n)}$$ Which test should be better to use here? I tried the D'Alembert's ratio test. $$lim_{nto infty} {frac{a^{ln(n)}}{a^{ln(n+1)}}}= lim_{nto infty}a^{ln{frac{n}{n+1}}}=a^0=1$$ So the test failed.



UPDATE: I observed that $${a^{ln(n)}}=e^{{ln(a)}{ln(n)}}={n^{ln(a)}}={{1over n^{ln({1over a})}} }$$Further by test of auxiliary series $$sum frac{1}{n^p}$$ we know that criterion for convergence is $p>1$.
Here $p=ln({1over a})$.Hence the series will be convergent when $ln(a)<-1$ i.e. when $0<a<1/e$, since $a>0$ is already given.Finally this solved my question.






sequences-and-series







share|cite|improve this question




















  • 1




    Which ones did you try already? Which ones do you know? What were the results? We aren't here to do your homework, you know.
    – 5xum
    Nov 14 '17 at 9:47












  • That's essentially the same as this question
    – Professor Vector
    Nov 14 '17 at 9:52






  • 1




    this converges if $$log(|a|)+1<0$$ is hold.
    – Dr. Sonnhard Graubner
    Nov 14 '17 at 9:58










  • @5xum I tried the D Alembert ratio test but it failed.
    – Shatabdi Sinha
    Nov 14 '17 at 10:07






  • 1




    Make yourself familiar with Raabe's test and the other ones on that page. But here, note that $a^{ln n} = expbigl((ln n)(ln a)bigr) = n^{ln a}$.
    – Daniel Fischer♦
    Nov 14 '17 at 10:26
















-2












-2








-2







For $a>0$, test the convergence of the infinite series: $$sum a^{ln(n)}$$ Which test should be better to use here? I tried the D'Alembert's ratio test. $$lim_{nto infty} {frac{a^{ln(n)}}{a^{ln(n+1)}}}= lim_{nto infty}a^{ln{frac{n}{n+1}}}=a^0=1$$ So the test failed.



UPDATE: I observed that $${a^{ln(n)}}=e^{{ln(a)}{ln(n)}}={n^{ln(a)}}={{1over n^{ln({1over a})}} }$$Further by test of auxiliary series $$sum frac{1}{n^p}$$ we know that criterion for convergence is $p>1$.
Here $p=ln({1over a})$.Hence the series will be convergent when $ln(a)<-1$ i.e. when $0<a<1/e$, since $a>0$ is already given.Finally this solved my question.






sequences-and-series







share|cite|improve this question















For $a>0$, test the convergence of the infinite series: $$sum a^{ln(n)}$$ Which test should be better to use here? I tried the D'Alembert's ratio test. $$lim_{nto infty} {frac{a^{ln(n)}}{a^{ln(n+1)}}}= lim_{nto infty}a^{ln{frac{n}{n+1}}}=a^0=1$$ So the test failed.



UPDATE: I observed that $${a^{ln(n)}}=e^{{ln(a)}{ln(n)}}={n^{ln(a)}}={{1over n^{ln({1over a})}} }$$Further by test of auxiliary series $$sum frac{1}{n^p}$$ we know that criterion for convergence is $p>1$.
Here $p=ln({1over a})$.Hence the series will be convergent when $ln(a)<-1$ i.e. when $0<a<1/e$, since $a>0$ is already given.Finally this solved my question.






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 14 '17 at 11:02

























asked Nov 14 '17 at 9:39









Shatabdi Sinha

17312




17312








  • 1




    Which ones did you try already? Which ones do you know? What were the results? We aren't here to do your homework, you know.
    – 5xum
    Nov 14 '17 at 9:47












  • That's essentially the same as this question
    – Professor Vector
    Nov 14 '17 at 9:52






  • 1




    this converges if $$log(|a|)+1<0$$ is hold.
    – Dr. Sonnhard Graubner
    Nov 14 '17 at 9:58










  • @5xum I tried the D Alembert ratio test but it failed.
    – Shatabdi Sinha
    Nov 14 '17 at 10:07






  • 1




    Make yourself familiar with Raabe's test and the other ones on that page. But here, note that $a^{ln n} = expbigl((ln n)(ln a)bigr) = n^{ln a}$.
    – Daniel Fischer♦
    Nov 14 '17 at 10:26
















  • 1




    Which ones did you try already? Which ones do you know? What were the results? We aren't here to do your homework, you know.
    – 5xum
    Nov 14 '17 at 9:47












  • That's essentially the same as this question
    – Professor Vector
    Nov 14 '17 at 9:52






  • 1




    this converges if $$log(|a|)+1<0$$ is hold.
    – Dr. Sonnhard Graubner
    Nov 14 '17 at 9:58










  • @5xum I tried the D Alembert ratio test but it failed.
    – Shatabdi Sinha
    Nov 14 '17 at 10:07






  • 1




    Make yourself familiar with Raabe's test and the other ones on that page. But here, note that $a^{ln n} = expbigl((ln n)(ln a)bigr) = n^{ln a}$.
    – Daniel Fischer♦
    Nov 14 '17 at 10:26










1




1




Which ones did you try already? Which ones do you know? What were the results? We aren't here to do your homework, you know.
– 5xum
Nov 14 '17 at 9:47






Which ones did you try already? Which ones do you know? What were the results? We aren't here to do your homework, you know.
– 5xum
Nov 14 '17 at 9:47














That's essentially the same as this question
– Professor Vector
Nov 14 '17 at 9:52




That's essentially the same as this question
– Professor Vector
Nov 14 '17 at 9:52




1




1




this converges if $$log(|a|)+1<0$$ is hold.
– Dr. Sonnhard Graubner
Nov 14 '17 at 9:58




this converges if $$log(|a|)+1<0$$ is hold.
– Dr. Sonnhard Graubner
Nov 14 '17 at 9:58












@5xum I tried the D Alembert ratio test but it failed.
– Shatabdi Sinha
Nov 14 '17 at 10:07




@5xum I tried the D Alembert ratio test but it failed.
– Shatabdi Sinha
Nov 14 '17 at 10:07




1




1




Make yourself familiar with Raabe's test and the other ones on that page. But here, note that $a^{ln n} = expbigl((ln n)(ln a)bigr) = n^{ln a}$.
– Daniel Fischer♦
Nov 14 '17 at 10:26






Make yourself familiar with Raabe's test and the other ones on that page. But here, note that $a^{ln n} = expbigl((ln n)(ln a)bigr) = n^{ln a}$.
– Daniel Fischer♦
Nov 14 '17 at 10:26












1 Answer
1






active

oldest

votes


















2














Correct me if wrong.



$a_n=a^{ln(n)}; $



Rewrite: $a =e^k$, then $k=ln(a).$



$a_n = [e^k]^{ln(n)} =e^{kln(n)}$



$a_n= n^k$, where $k= ln(a) in mathbb{R}.$



$sum a^{ln(n)}:$



1) $-1 le k$, divergent.



4) $klt -1$, convergent.






share|cite|improve this answer























  • Yes that is right. I used the same approach. Please see my update in the question.
    – Shatabdi Sinha
    Nov 14 '17 at 11:09










  • Shatabdi. Thanks for your comment.We are on the same track.:))))
    – Peter Szilas
    Nov 14 '17 at 11:16











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2














Correct me if wrong.



$a_n=a^{ln(n)}; $



Rewrite: $a =e^k$, then $k=ln(a).$



$a_n = [e^k]^{ln(n)} =e^{kln(n)}$



$a_n= n^k$, where $k= ln(a) in mathbb{R}.$



$sum a^{ln(n)}:$



1) $-1 le k$, divergent.



4) $klt -1$, convergent.






share|cite|improve this answer























  • Yes that is right. I used the same approach. Please see my update in the question.
    – Shatabdi Sinha
    Nov 14 '17 at 11:09










  • Shatabdi. Thanks for your comment.We are on the same track.:))))
    – Peter Szilas
    Nov 14 '17 at 11:16
















2














Correct me if wrong.



$a_n=a^{ln(n)}; $



Rewrite: $a =e^k$, then $k=ln(a).$



$a_n = [e^k]^{ln(n)} =e^{kln(n)}$



$a_n= n^k$, where $k= ln(a) in mathbb{R}.$



$sum a^{ln(n)}:$



1) $-1 le k$, divergent.



4) $klt -1$, convergent.






share|cite|improve this answer























  • Yes that is right. I used the same approach. Please see my update in the question.
    – Shatabdi Sinha
    Nov 14 '17 at 11:09










  • Shatabdi. Thanks for your comment.We are on the same track.:))))
    – Peter Szilas
    Nov 14 '17 at 11:16














2












2








2






Correct me if wrong.



$a_n=a^{ln(n)}; $



Rewrite: $a =e^k$, then $k=ln(a).$



$a_n = [e^k]^{ln(n)} =e^{kln(n)}$



$a_n= n^k$, where $k= ln(a) in mathbb{R}.$



$sum a^{ln(n)}:$



1) $-1 le k$, divergent.



4) $klt -1$, convergent.






share|cite|improve this answer














Correct me if wrong.



$a_n=a^{ln(n)}; $



Rewrite: $a =e^k$, then $k=ln(a).$



$a_n = [e^k]^{ln(n)} =e^{kln(n)}$



$a_n= n^k$, where $k= ln(a) in mathbb{R}.$



$sum a^{ln(n)}:$



1) $-1 le k$, divergent.



4) $klt -1$, convergent.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 14 '17 at 11:06


























community wiki





2 revs
Peter Szilas













  • Yes that is right. I used the same approach. Please see my update in the question.
    – Shatabdi Sinha
    Nov 14 '17 at 11:09










  • Shatabdi. Thanks for your comment.We are on the same track.:))))
    – Peter Szilas
    Nov 14 '17 at 11:16


















  • Yes that is right. I used the same approach. Please see my update in the question.
    – Shatabdi Sinha
    Nov 14 '17 at 11:09










  • Shatabdi. Thanks for your comment.We are on the same track.:))))
    – Peter Szilas
    Nov 14 '17 at 11:16
















Yes that is right. I used the same approach. Please see my update in the question.
– Shatabdi Sinha
Nov 14 '17 at 11:09




Yes that is right. I used the same approach. Please see my update in the question.
– Shatabdi Sinha
Nov 14 '17 at 11:09












Shatabdi. Thanks for your comment.We are on the same track.:))))
– Peter Szilas
Nov 14 '17 at 11:16




Shatabdi. Thanks for your comment.We are on the same track.:))))
– Peter Szilas
Nov 14 '17 at 11:16


















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