Explicație pas cu pas:
Observam ca termenii sumei din paranteza formeaza o progresie [tex](b_n)_{n}}[/tex]geometrica cu primul termen:
[tex]b_1=\frac{1}{e}[/tex]
si ratia
[tex]q=\frac{b_2}{b_1}=\frac{\frac{1}{e^2}}{\frac{1}{e}}=\frac{1}{e^2}\cdot e=\frac{1}{e}[/tex].
Calculam suma primilor n termeni ai acestei progresii:
[tex]S_n=b_1 \cdot \frac{q^n-1}{q-1}=\frac{1}{e}\cdot \frac{(\frac{1}{e}^n)-1}{\frac{1}{e}-1}=\frac{1}{e}\cdot\frac{\frac{1}{e^n}-\frac{e^n}{e^n}}{\frac{1}{e}-\frac{e}{e}}=\frac{1}{e}\cdot\frac{\frac{1-e^n}{e^n}}{\frac{1-e}{e}}=\frac{1}{e}\cdot\frac{1-e^n}{e^n}\cdot\frac{e}{1-e}=\frac{1-e^n}{e^n(1-e)}[/tex]
Calculam limita acestei sume:
[tex]\lim_{n \to \infty} \frac{1-e^n}{e^n(1-e)}= \lim_{n \to \infty}\frac{e^n(\frac{1}{e^n}-1)}{e^n(1-e)}=\lim_{n \to \infty}\frac{\frac{1}{e^n}-1}{1-e}=\frac{\frac{1}{\infty}-1}{1-e}=\frac{0-1}{1-e}=\frac{-1}{1-e}=\frac{1}{-(1-e)}=\frac{1}{-1+e}=\frac{1}{e-1}[/tex]
