Question

Demonstrati ca:
$\lim_{n \to \infty} (\frac{1}{e} +\frac{1}{e^{2} } +...+\frac{1}{e^{n} } )$$=\frac{1}{e-1}$

Answer 1

Răspuns:

Ai raspunsul atasat

Explicație pas cu pas:

Answer 2

Explicație pas cu pas:

Observam ca termenii sumei din paranteza formeaza o progresie $(b_n)_{n}}$geometrica cu primul termen:

$b_1=\frac{1}{e}$

si ratia

$q=\frac{b_2}{b_1}=\frac{\frac{1}{e^2}}{\frac{1}{e}}=\frac{1}{e^2}\cdot e=\frac{1}{e}$.

Calculam suma primilor n termeni ai acestei progresii:

$S_n=b_1 \cdot \frac{q^n-1}{q-1}=\frac{1}{e}\cdot \frac{(\frac{1}{e}^n)-1}{\frac{1}{e}-1}=\frac{1}{e}\cdot\frac{\frac{1}{e^n}-\frac{e^n}{e^n}}{\frac{1}{e}-\frac{e}{e}}=\frac{1}{e}\cdot\frac{\frac{1-e^n}{e^n}}{\frac{1-e}{e}}=\frac{1}{e}\cdot\frac{1-e^n}{e^n}\cdot\frac{e}{1-e}=\frac{1-e^n}{e^n(1-e)}$

Calculam limita acestei sume:

$\lim_{n \to \infty} \frac{1-e^n}{e^n(1-e)}= \lim_{n \to \infty}\frac{e^n(\frac{1}{e^n}-1)}{e^n(1-e)}=\lim_{n \to \infty}\frac{\frac{1}{e^n}-1}{1-e}=\frac{\frac{1}{\infty}-1}{1-e}=\frac{0-1}{1-e}=\frac{-1}{1-e}=\frac{1}{-(1-e)}=\frac{1}{-1+e}=\frac{1}{e-1}$