Question

Evaluate $\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{2^r}$

• A. 0
• B. 1
• C. $\infty$
• D. $\frac{1}{2}$
• E. 2

Answer 1

Answer: (B)

1

$\sum _{r=1}^n\frac{1}{2^r}=\frac{\frac{1}{2}\{1-{\left(\frac{1}{2}\right)}^n\}}{\{1-\frac{1}{2}\}}=1-{\left(\frac{1}{2}\right)}^n$
which tends to $1-0=1$ as $n\to \infty$.
$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{2^2}=1$
Hence, option 'B' is correct.