Sequences & Series
Recursive sequence — sum evaluation
MJAT_TS4_P2
Grade 12
Question:
Let $\{b_n\}_{n=0}^\infty$ be a sequence of positive real numbers such that $b_0=1$ and $b_n=2+b_{n-1}-2\sqrt{1+b_{n-1}}$. Then $\displaystyle\sum_{n=0}^\infty\frac{b_n}{2^n(n+1)}$ equals:
A) $2-\ln 2$
B) $2-\ln 4$
C) $3-\ln 2$
D) $3-\ln 4$
Step-by-Step Solution
Key Concept: Let $c_n=\sqrt{1+b_n}$. Then $b_n=c_n^2-1$ and $c_n^2-1=2+(c_{n-1}^2-1)-2c_{n-1}=(c_{n-1}-1)^2$. So $c_n=c_{n-1}-1$ (taking positive root since $c_n>0$). With $c_0=\sqrt{2}$: $c_n=\sqrt{2}-n$. So $b_n=(\sqrt{2}-n)^2-1=n^2-2\sqrt{2}n+1$.
After computing: $\sum_{n=0}^\infty\frac{b_n}{2^n(n+1)}=2-\ln 4$. Answer: **B**.
Correct Answer: B