Răspuns:
Explicație pas cu pas:
[tex]\displaystyle a_n=\lim_{x\to 0}(1-\sin nx)^{\frac{1}{x}}=\lim_{x\to 0}{[(1-\sin nx)^{\frac{-1}{\sin nx}}]}^{\frac{-\sin nx}{x}}=\lim_{x\to 0}e^{-\frac{\sin nx}{x}}=\\=e^{-n}=\dfrac{1}{e^n}\\\lim_{n\to\infty}(a_1+a_2+\ldots+a_n)=\lim_{n\to\infty}\left(\dfrac{1}{e}+\dfrac{1}{e^2}+\dfrac{1}{e^3}+\ldots+\dfrac{1}{e^n}\right)=\\=\lim_{n\to\infty}\dfrac{1}{e}\cdot \dfrac{1-\left(\frac{1}{e}\right)^n}{1-\frac{1}{e}}=\dfrac{e}{e(e-1)}=\dfrac{1}{e-1}\\\texttt{Prin urmare raspunsul este b)}.[/tex]
