Question

$lim_{n→∞} \frac1{2^n}(\frac1{\sqrt{1-1/2^n}} +\frac1{\sqrt{1-\frac2{2^n}}}+\frac1{\sqrt{1-\frac3{2^n}}}+...+\frac1{\sqrt{1-\frac{2^n-1}{2^n}}})$ is equal to

• A. $\frac1{2}$
• B. 1
• C. 2
• D. -2

Answer 1

Answer: (C)

2

Answer 2

$I =lim_{n→∞} \frac1{2^n}(\frac1{\sqrt{1-1/2^n}} +\frac1{\sqrt{1-\frac2{2^n}}}+\frac1{\sqrt{1-\frac3{2^n}}}+...+\frac1{\sqrt{1-\frac{2^n-1}{2^n}}})$

Let $2^n=t$ and if$n→∞$ then $t→∞$

$I= lim_{n→∞} \frac1{t} (\overset{t=1}{\underset{r=1}\sum} \frac1{√1-\frac{r}{t}})$

$l= ∫_0 ^1 \frac{dx}{√1-x}=∫_0^1 \frac {dx}{\sqrt {x}}$ $\bf{∫_0^a f(x)dx=∫_0^a f(a-x)dx }$

$=[2x^{\frac1{2}}]^1_0$

$=2$