[tex]\displaystyle\sum_{n=1}^\infty\,\,\frac{ \left(\displaystyle\sum_{j=1}^n\,\,2^j\right)^{\alpha+1}}{ \left(\displaystyle\sum_{k=1}^n\,\,\frac{1}{k}\right)^{\frac{\alpha}{2} }} ,\qquad\alpha\in\mathbb{R}[/tex]
La serie è fortunatamente a termini positivi; consideriamo i vari termini del termine generale della serie:
[tex]\left(\displaystyle\sum_{j=1}^n\,\,2^j\right)^{\alpha+1}:[/tex]
si tratta della somma di [tex]n[/tex] termini in progressione geometrica di ragione [tex]2;[/tex]come noto:
[tex]\displaystyle\sum_{k=0}^n\,\,x^k = \frac{1-x^{n+1}}{1-x};[/tex]
allora nel nostro caso avremo:
[tex]\displaystyle\sum_{j=1}^n\,\,2^j=2 +2^2+2^3+\cdots+2^n =[/tex] [tex]\displaystyle2(1 +2 +2^2+\cdots+2^{n-1}) =[/tex] [tex]\displaystyle2 \sum_{j=0}^{n-1}2^{j}
=2\left(\frac{1-2^{n-1+1}}{1-2}\right)=-2 ( 1-2^{n } ) =2^{n+1}-2[/tex]
e dunque:
[tex]\left(\displaystyle\sum_{j=1}^n\,\,2^j\right)^{\alpha+1}=\left( 2^{n+1}-2\right)^{\alpha+1}[/tex]
[tex]\displaystyle\left(\displaystyle\sum_{k=1}^n\,\,\frac{1}{k}\right)^{\frac{\alpha}{2} }[/tex]
ricordando l'apprissimazione asintotica della serie armonica, si ha:
[tex]\displaystyle\sum_{k=1}^n\,\,\frac{1}{k}\sim \ln n\Rightarrow \left( \ln n\right)^{\frac{\alpha}{2}}[/tex]
Fatte queste presmesse, la serie data diventa:
[tex]\displaystyle\sum_{n=1}^\infty\,\,\frac{ \left(\displaystyle\sum_{j=1}^n\,\,2^j\right)^{\alpha+1}}{ \left(\displaystyle\sum_{k=1}^n\,\,\frac{1}{k}\right)^{\frac{\alpha}{2} }}[/tex] [tex]\displaystyle \sim \displaystyle\sum_{n=1}^\infty\,\,\frac{\left( 2^{n+1}-2\right)^{\alpha+1}}{\left( \ln n\right)^{\frac{\alpha}{2}}}\stackrel{Ratio}{\Longrightarrow}[/tex][tex]\displaystyle\lim_{n\to+\infty} \frac{\left( 2^{n+2}-2\right)^{\alpha+1}}{\left( \ln (n+1)\right)^{\frac{\alpha}{2}}}\cdot \frac{\left( \ln n\right)^{\frac{\alpha}{2}}}{\left( 2^{n+1}-2\right)^{\alpha+1}}[/tex] [tex]\displaystyle=\lim_{n\to+\infty} \frac{\left( 2^ n\cdot 2^2 -2\right)^{\alpha}\cdot \left( 2^ n\cdot 2^2 -2\right) }{\left( \ln (n+1)\right)^{\frac{\alpha}{2}}}\cdot \frac{\left( \ln n\right)^{\frac{\alpha}{2}}}{\left( 2^ n\cdot2-2\right)^{\alpha}\cdot \left( 2^ n\cdot2-2\right) }[/tex] [tex]\displaystyle=\lim_{n\to+\infty} \frac{2^{\alpha+1}\left( 2^ n\cdot 2 -1\right)^{\alpha}\cdot \left( 2^ n\cdot 2 -1\right) }{2^{\alpha+1 }\left( 2^ n -1\right)^{\alpha}\cdot \left( 2^ n -1\right)}\cdot \left( \frac{ \ln n }{ \ln (n+1)}\right)^{\frac{\alpha}{2}}[/tex]
[tex]\displaystyle\sim\lim_{n\to+\infty} \frac{ \left( 2^ n\cdot 2 \right)^{\alpha}\cdot \left( 2^ n\cdot 2 \right) }{ \left( 2^ n \right)^{\alpha}\cdot \left( 2^ n \right)}\cdot 1=\lim_{n\to+\infty} \frac{ 2 ^{n\alpha}\cdot 2 ^{\alpha} \cdot 2^ n\cdot 2 }{ 2^{n\alpha}\cdot2^ n }=2^{\alpha+1}[/tex]
a questo punto, affinche il criterio del rapporto sia efficace deve essere:
[tex]\displaystyle2^{\alpha+1}<1 \Rightarrow \alpha+1<0 \Rightarrow \alpha<-1[/tex]
Si conclude dunque che la serie data risulta convergente per [tex]\alpha<-1.[/tex]
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speriamo bene....
