Question

Suppose that we have the following sequence :

Answer 1

$a_n=\dfrac12a_{n-1}$
$a_n=\dfrac1{2^2}a_{n-2}$
$a_n=\dfrac1{2^3}a_{n-3}$
$a_n=\cdots=\dfrac1{2^{n-1}}a_1$
$a_n=\dfrac1{2^{n-1}}$

$b_n=b_{n-1}+\dfrac1{2^{n-1}}$
$b_n=b_{n-2}+\dfrac1{2^{n-1}}+\dfrac1{2^{n-2}}$
$b_n=b_{n-3}+\dfrac1{2^{n-1}}+\dfrac1{2^{n-2}}+\dfrac1{2^{n-3}}$
$b_n=\cdots=b_1+\dfrac1{2^{n-1}}+\dfrac1{2^{n-2}}+\cdots+\dfrac12$
$b_n=a_1+\displaystyle\sum_{k=1}^{n-1}\frac1{2^{n-k}}$
$b_n=1+\displaystyle\sum_{k=1}^{n-1}\frac1{2^{n-k}}$
$b_n=\displaystyle\sum_{k=1}^n\frac1{2^{n-k}}$
$b_n=\displaystyle\frac1{2^n}\underbrace{\sum_{k=1}^n2^k}_{S_n}$

$S_n=1+2+2^2+\cdots+2^{n-1}+2^n$
$\implies2S_n=2+2^2+2^3+\cdots+2^n+2^{n+1}$
$\implies S_n-2S_n=-S_n=1-2^{n+1}$
$\implies S_n=2^{n+1}-1$

$b_n=\dfrac{2^{n+1}-1}{2^n}=2-\dfrac1{2^n}$

$\implies b_{50}=2-\dfrac1{2^{50}}\approx1.99999999999999911182158$

$\implies b_{10^6}=2-\dfrac1{2^{10^6}}\approx2.00000000000000000000000$