Răspuns:
limita (1/3)^n+1 tinde la zero
Explicație pas cu pas:
In paranteze avem progresii geometrice, in prima ratia q1=1/3, in a doua ratia q2=1/2 si primul termen la ambele este 1, iar numarul de termeni este n+1.
Sumele din paranteze:
[tex]S1=a1*\frac{1-q^{n+1}}{1-q}=1*\frac{1-(\frac{1}{3})^{n+1} }{1-\frac{1}{3} }= \frac{1-(\frac{1}{3})^{n+1} }{\frac{2}{3} }=\frac{3}{2}*( 1-(\frac{1}{3})^{n+1}).\\S2=a1*\frac{1-q^{n+1}}{1-q}=1*\frac{1-(\frac{1}{2})^{n+1} }{1-\frac{1}{2} }= \frac{1-(\frac{1}{2})^{n+1} }{\frac{1}{2} }=\frac{2}{1}*( 1-(\frac{1}{2})^{n+1})[\tex]\\a_{n}=S1:S2=\frac{3}{2}*( 1-(\frac{1}{3})^{n+1}):[\frac{2}{1}*( 1-(\frac{1}{2})^{n+1})]\\ \lim_{n \to \infty} a_n =\frac{3}{2}*(1-0):[\frac{2}{1}*(1-0)]=\frac{3}{2}:\frac{2}{1}=\frac{3}{2}*\frac{1}{2}=\frac{3}{4}.[/tex]
