Question

Suppose that we have the following sequence :

a_1=1
a_n=1/2 a_(n-1) , for n>1

We define a new sequence as such :
b_1=a_1
b_n=b_(n-1)+a_n

Find the 50th term of the sequence b_n.
Find the 1000000th term of the sequence b_n.

Answer 1

`a_n=\frac12a_(n-1)`

`a_n=\frac1{2^2}a_(n-2)`

`a_n=\frac1{2^3}a_(n-3)`

`a_n=\cdots=\frac1{2^(n-1)}a_1`

`a_n=\frac1{2^(n-1)}`


`b_n=b_(n-1)+\frac1{2^(n-1)}`

`b_n=b_(n-2)+\frac1{2^(n-1)}+\frac1{2^(n-2)}`

`b_n=b_(n-3)+\frac1{2^(n-1)}+\frac1{2^(n-2)}+\frac1{2^(n-3)}`

`b_n=\cdots=b_1+\frac1{2^(n-1)}+\frac1{2^(n-2)}+\cdots+\frac12`

`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=(2^(n+1)-1)/(2^n)=2-\frac1{2^n}`


`\implies b_(50)=2-\frac1{2^(50)}\approx1.99999999999999911182158`


`\implies b_(10^6)=2-\frac1{2^(10^6)}\approx2.00000000000000000000000`