Proving divergence of series












1












$begingroup$


I am attempting to establish divergence of the following series at $x=1/e$
$$sum_{n=1}^infty x^{1+frac{1}{2}+frac{1}{3}+cdots+frac{1}{n}}$$
Most of my ideas have revolved around the comparison test, but I haven't been able to find a divergent series which is less than my original series. Any help would be appreciated.






sequences-and-series convergence







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  • 1




    $begingroup$
    Set $x=e^{-1}$ and $a_n=x^{1+ldots+1/n}$. Prove that $|ln{(n+1)a_{n+1}}-ln{na_n}|$ has a finite sum, thus $a_n sim c/n$ for some $c >0$.
    $endgroup$
    – Mindlack
    Jan 16 at 15:42
















1












$begingroup$


I am attempting to establish divergence of the following series at $x=1/e$
$$sum_{n=1}^infty x^{1+frac{1}{2}+frac{1}{3}+cdots+frac{1}{n}}$$
Most of my ideas have revolved around the comparison test, but I haven't been able to find a divergent series which is less than my original series. Any help would be appreciated.






sequences-and-series convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Set $x=e^{-1}$ and $a_n=x^{1+ldots+1/n}$. Prove that $|ln{(n+1)a_{n+1}}-ln{na_n}|$ has a finite sum, thus $a_n sim c/n$ for some $c >0$.
    $endgroup$
    – Mindlack
    Jan 16 at 15:42














1












1








1





$begingroup$


I am attempting to establish divergence of the following series at $x=1/e$
$$sum_{n=1}^infty x^{1+frac{1}{2}+frac{1}{3}+cdots+frac{1}{n}}$$
Most of my ideas have revolved around the comparison test, but I haven't been able to find a divergent series which is less than my original series. Any help would be appreciated.






sequences-and-series convergence







share|cite|improve this question









$endgroup$




I am attempting to establish divergence of the following series at $x=1/e$
$$sum_{n=1}^infty x^{1+frac{1}{2}+frac{1}{3}+cdots+frac{1}{n}}$$
Most of my ideas have revolved around the comparison test, but I haven't been able to find a divergent series which is less than my original series. Any help would be appreciated.






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 15:31









s0ulr3aper07s0ulr3aper07

32110




32110








  • 1




    $begingroup$
    Set $x=e^{-1}$ and $a_n=x^{1+ldots+1/n}$. Prove that $|ln{(n+1)a_{n+1}}-ln{na_n}|$ has a finite sum, thus $a_n sim c/n$ for some $c >0$.
    $endgroup$
    – Mindlack
    Jan 16 at 15:42














  • 1




    $begingroup$
    Set $x=e^{-1}$ and $a_n=x^{1+ldots+1/n}$. Prove that $|ln{(n+1)a_{n+1}}-ln{na_n}|$ has a finite sum, thus $a_n sim c/n$ for some $c >0$.
    $endgroup$
    – Mindlack
    Jan 16 at 15:42








1




1




$begingroup$
Set $x=e^{-1}$ and $a_n=x^{1+ldots+1/n}$. Prove that $|ln{(n+1)a_{n+1}}-ln{na_n}|$ has a finite sum, thus $a_n sim c/n$ for some $c >0$.
$endgroup$
– Mindlack
Jan 16 at 15:42




$begingroup$
Set $x=e^{-1}$ and $a_n=x^{1+ldots+1/n}$. Prove that $|ln{(n+1)a_{n+1}}-ln{na_n}|$ has a finite sum, thus $a_n sim c/n$ for some $c >0$.
$endgroup$
– Mindlack
Jan 16 at 15:42










1 Answer
1






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oldest

votes


















1












$begingroup$

Render



$(1/1)+(1/2)+(1/3)+...+(1/n)<1+int_1^n(1/t)dt=1+ln(n)$



due to $f(t)=1/t$ being monotonically decreasing for positive $t$. Then



$(1/e)^{(1/1)+(1/2)+(1/3)+...+(1/n)}>(1/e)^{1+ln(n)}=(1/(en))$



and your series compares with the harmonic series.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Typo in the final identity.
    $endgroup$
    – Yves Daoust
    Jan 16 at 15:58






  • 1




    $begingroup$
    @Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 17:17






  • 1




    $begingroup$
    A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 17:57










  • $begingroup$
    @Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 18:51












  • $begingroup$
    You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 20:42













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Render



$(1/1)+(1/2)+(1/3)+...+(1/n)<1+int_1^n(1/t)dt=1+ln(n)$



due to $f(t)=1/t$ being monotonically decreasing for positive $t$. Then



$(1/e)^{(1/1)+(1/2)+(1/3)+...+(1/n)}>(1/e)^{1+ln(n)}=(1/(en))$



and your series compares with the harmonic series.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Typo in the final identity.
    $endgroup$
    – Yves Daoust
    Jan 16 at 15:58






  • 1




    $begingroup$
    @Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 17:17






  • 1




    $begingroup$
    A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 17:57










  • $begingroup$
    @Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 18:51












  • $begingroup$
    You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 20:42


















1












$begingroup$

Render



$(1/1)+(1/2)+(1/3)+...+(1/n)<1+int_1^n(1/t)dt=1+ln(n)$



due to $f(t)=1/t$ being monotonically decreasing for positive $t$. Then



$(1/e)^{(1/1)+(1/2)+(1/3)+...+(1/n)}>(1/e)^{1+ln(n)}=(1/(en))$



and your series compares with the harmonic series.






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Typo in the final identity.
    $endgroup$
    – Yves Daoust
    Jan 16 at 15:58






  • 1




    $begingroup$
    @Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 17:17






  • 1




    $begingroup$
    A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 17:57










  • $begingroup$
    @Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 18:51












  • $begingroup$
    You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 20:42
















1












1








1





$begingroup$

Render



$(1/1)+(1/2)+(1/3)+...+(1/n)<1+int_1^n(1/t)dt=1+ln(n)$



due to $f(t)=1/t$ being monotonically decreasing for positive $t$. Then



$(1/e)^{(1/1)+(1/2)+(1/3)+...+(1/n)}>(1/e)^{1+ln(n)}=(1/(en))$



and your series compares with the harmonic series.






share|cite|improve this answer











$endgroup$



Render



$(1/1)+(1/2)+(1/3)+...+(1/n)<1+int_1^n(1/t)dt=1+ln(n)$



due to $f(t)=1/t$ being monotonically decreasing for positive $t$. Then



$(1/e)^{(1/1)+(1/2)+(1/3)+...+(1/n)}>(1/e)^{1+ln(n)}=(1/(en))$



and your series compares with the harmonic series.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 17:50

























answered Jan 16 at 15:50









Oscar LanziOscar Lanzi

12.7k12136




12.7k12136








  • 3




    $begingroup$
    Typo in the final identity.
    $endgroup$
    – Yves Daoust
    Jan 16 at 15:58






  • 1




    $begingroup$
    @Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 17:17






  • 1




    $begingroup$
    A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 17:57










  • $begingroup$
    @Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 18:51












  • $begingroup$
    You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 20:42
















  • 3




    $begingroup$
    Typo in the final identity.
    $endgroup$
    – Yves Daoust
    Jan 16 at 15:58






  • 1




    $begingroup$
    @Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 17:17






  • 1




    $begingroup$
    A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 17:57










  • $begingroup$
    @Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
    $endgroup$
    – s0ulr3aper07
    Jan 16 at 18:51












  • $begingroup$
    You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
    $endgroup$
    – Oscar Lanzi
    Jan 16 at 20:42










3




3




$begingroup$
Typo in the final identity.
$endgroup$
– Yves Daoust
Jan 16 at 15:58




$begingroup$
Typo in the final identity.
$endgroup$
– Yves Daoust
Jan 16 at 15:58




1




1




$begingroup$
@Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
$endgroup$
– s0ulr3aper07
Jan 16 at 17:17




$begingroup$
@Oscar The use of that precise integral to arrive at the inequality is quite an elegant solution. I wish I could come up with such ideas. Thanks a lot!
$endgroup$
– s0ulr3aper07
Jan 16 at 17:17




1




1




$begingroup$
A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
$endgroup$
– Oscar Lanzi
Jan 16 at 17:57




$begingroup$
A secondary exercise: Use the integration of the reciprocal function to render $(sum)>ln(n)$, noting the reversed inequality sign. Show that if we then make the base any positive number less than $1/e$, the series becomes convergent.
$endgroup$
– Oscar Lanzi
Jan 16 at 17:57












$begingroup$
@Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
$endgroup$
– s0ulr3aper07
Jan 16 at 18:51






$begingroup$
@Oscar, If $x<frac{1}{e}$, then $x^{sum}<frac{1}{ne^{1/n}}$ (using above mentioned inequality). Now using Ratio Test on $sumfrac{1}{ne^{1/n}}$, we can show that it converges, which implies that $sum x^{sum}$ also converges. Am I correct?
$endgroup$
– s0ulr3aper07
Jan 16 at 18:51














$begingroup$
You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
$endgroup$
– Oscar Lanzi
Jan 16 at 20:42






$begingroup$
You should get a comparison term (proportional to) $n^{ln(x)}$ where $x$ is the base. A big hint: exponentiation has a quasi- commutative law, $a^{ln(b)}=b^{ln(a)}$.
$endgroup$
– Oscar Lanzi
Jan 16 at 20:42




















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