Réponse :
Bonjour,
Explications étape par étape :
[tex]\left\{\begin{array}{ccc}u_0&=&1\\u_{n+1}&=&exp(\dfrac{1}{2^n} )*u_n\\\end {array} \right.[/tex]
[tex]u_0=1\\u_1=exp(\dfrac{1}{2^0})*1=e\\u_2=exp(\dfrac{1}{2^1})*e=exp(\dfrac{3}{2} )\\\\v_0=ln(u_0)=ln(1)=0\\v_1=ln(u_1)=ln(exp(1))=1\\v_2=ln(u_1)=ln(exp(\frac{3}{2} ))=\dfrac{3}{2} \\[/tex]
Variation de la suite v:
[tex]v_{n+1}-v_n=ln(exp(\frac{1}{2^n}))*u_n-ln(u_n)=ln( exp(\frac{1}{2^n}))\\=\dfrac{1}{2^n} > \ 0\\[/tex]
La suite v est croissante.
[tex]v_0=0\\v_1-v_0=v_1=\dfrac{1}{2^0} \\v_2-v_1=\dfrac{1}{2^1}\\v_3-v_2=\dfrac{1}{2^2}\\...\\v_{n+1}-v_n=\dfrac{1}{2^n}\\\\\Longrightarrow\ v_{n+1}=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2^n}\\\\\Longrightarrow\ v_{n}=\displaystyle \sum_{k=0}^{n-1}\frac{1}{2^k} =1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2^{n-1}}\\=\dfrac{1-\frac{1}{2^n}}{1-\frac{1}{2}} \\\boxed{=2-\dfrac{1}{2^{n-1}}}\\[/tex]
[tex]\displaystyle \lim_{n \to \infty} (2-\dfrac{1}{2^{n-1}}})= 2=lim\ v_n\\\\\displaystyle \lim_{n \to \infty} u_n =e^2\\[/tex]
