Va rog sa ma ajutati cu urmatorul exercitiu! Se considera sirul
Sa se demonstreze ca
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Integram
prin parti:
dx=\int_0^1&space;\left&space;(&space;\frac{x^{n+1}}{n+1}&space;\right&space;)'\cdot\ln(x+1)dx=\frac1{n+1}\int_0^1(x^{n+1})'\ln(x+1)dx=\frac1{n+1}(x^{n+1}\ln(x+1)|_0^1-\int_0^1x^{n+1}(\ln(x+1))'dx)=\frac1{n+1}(1^{n+1}\ln(1+1)-0^{n+1}\ln(0+1)-\int_0^1\frac{x^{n+1}}{x+1}dx)=\frac1{n+1}(\ln2-\int_0^1\frac{x^{n+1}}{x+1}dx))
Inlocuim noua formula a lui
in limita:
![\lim_{n\to\infty}(n+2)[(n+1)a_n-\ln2]=\lim_{n\to\infty}(n+2)[(n+1)\cdot\frac1{n+1}(\ln2-\int_0^1\frac{x^{n+1}}{x+1}dx)-\ln2]=\lim_{n\to\infty}(n+2)[\ln2-\int_0^1\frac{x^{n+1}}{x+1}dx-\ln2]=\lim_{n\to\infty}(n+2)[-\int_0^1\frac{x^{n+1}}{x+1}dx]=-\lim_{n\to\infty}(n+2)\int_0^1\frac{x^{n+1}}{x+1}dx=-\lim_{n\to\infty}\int_0^1\frac{(n+2)x^{n+1}}{x+1}dx=-\lim_{n\to\infty}\int_0^1\frac{(x^{n+2})'}{x+1}dx=-\lim_{n\to\infty}((\frac{x^{n+2}}{x+1})|_0^1-\int_0^1x^{n+2}\cdot(\frac1{x+1})'dx)=-\lim_{n\to\infty}(\frac{1^{n+2}}{1+1}-\frac{0^{n+2}}{0+1}-\int_0^1x^{n+2}\cdot(\frac{-1}{(x+1)^2})dx)=-\lim_{n\to\infty}(\frac12+\int_0^1\frac{x^{n+2}}{(x+1)^2}dx)=-\frac12-\lim_{n\to\infty}\int_0^1\frac{x^{n+2}}{(x+1)^2}dx](http://latex.codecogs.com/gif.latex?\lim_{n\to\infty}(n+2)[(n+1)a_n-\ln2]=\lim_{n\to\infty}(n+2)[(n+1)\cdot\frac1{n+1}(\ln2-\int_0^1\frac{x^{n+1}}{x+1}dx)-\ln2]=\lim_{n\to\infty}(n+2)[\ln2-\int_0^1\frac{x^{n+1}}{x+1}dx-\ln2]=\lim_{n\to\infty}(n+2)[-\int_0^1\frac{x^{n+1}}{x+1}dx]=-\lim_{n\to\infty}(n+2)\int_0^1\frac{x^{n+1}}{x+1}dx=-\lim_{n\to\infty}\int_0^1\frac{(n+2)x^{n+1}}{x+1}dx=-\lim_{n\to\infty}\int_0^1\frac{(x^{n+2})'}{x+1}dx=-\lim_{n\to\infty}((\frac{x^{n+2}}{x+1})|_0^1-\int_0^1x^{n+2}\cdot(\frac1{x+1})'dx)=-\lim_{n\to\infty}(\frac{1^{n+2}}{1+1}-\frac{0^{n+2}}{0+1}-\int_0^1x^{n+2}\cdot(\frac{-1}{(x+1)^2})dx)=-\lim_{n\to\infty}(\frac12+\int_0^1\frac{x^{n+2}}{(x+1)^2}dx)=-\frac12-\lim_{n\to\infty}\int_0^1\frac{x^{n+2}}{(x+1)^2}dx)
Mai ramane sa calculam limita acelei integrale. Deoarece functia pe care o integram este continua, din teorema de medie stim ca exista
astfel incat
.
^2}dx=\lim_{n\to\infty}\frac{c^{n+2}}{(c+1)^2}=0)
Limita este 0 deoarece c-ul se afla intre 0 si 1, deci limita numitorului este 0, iar numaratorul este doar 0 constanta. Daca nu cunosti teorema de medie pe care am folosit-o mai sus, pot cauta o solutie alternativa.
Continuand limita de mai sus, avem
.
Multumesc mult!