Question

$\displaystyle\lim _{n \rightarrow \infty} \frac{1}{2^n}\left(\frac{1}{\sqrt{1-\frac{1}{2^n}}}+\frac{1}{\sqrt{1-\frac{2}{2^n}}}+\frac{1}{\sqrt{1-\frac{3}{2^n}}}+\ldots +\frac{1}{\sqrt{1-\frac{2^n-1}{2^n}}}\right)$ is equal to

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

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate the following limit:

\(\lim _{n \rightarrow \infty} \frac{1}{2^n}\left(\frac{1}{\sqrt{1-\frac{1}{2^n}}}+\frac{1}{\sqrt{1-\frac{2}{2^n}}}+\ldots +\frac{1}{\sqrt{1-\frac{2^n-1}{2^n}}}\right)\)

This expression can be interpreted as the Riemann sum approximation to an integral as \(n\) approaches infinity. We will derive the solution by converting this sum into an integral.

Notice that the sum inside the limit is a Riemann sum for the interval \([0, 1]\) with each subinterval of length \(\frac{1}{2^n}\). Therefore, the function inside the sum can be represented as:

\(\frac{1}{\sqrt{1-\frac{k}{2^n}}} \approx f\left(\frac{k}{2^n}\right)\)

where \(f(x) = \frac{1}{\sqrt{1-x}}\).

The Riemann sum is expressed as:

\(\frac{1}{2^n} \sum_{k=1}^{2^n - 1} f\left(\frac{k}{2^n}\right)\)

As \(n \to \infty\), this sum converges to the integral:

\(\int_0^1 \frac{1}{\sqrt{1-x}} \, dx\)

To solve the integral \(\int_0^1 \frac{1}{\sqrt{1-x}} \, dx\), substitute \(u = 1-x\), which gives \(du = -dx\) or \(dx = -du\). The limits of integration change accordingly:

• When \(x = 0\), \(u = 1\)
• When \(x = 1\), \(u = 0\)

The integral \(\int u^{-\frac{1}{2}} \, du\) is \(\frac{u^{\frac{1}{2}}}{\frac{1}{2}} = 2u^{\frac{1}{2}}\). Evaluating from 0 to 1:

\(\left[2u^{\frac{1}{2}}\right]_0^1 = 2 \times (1^{\frac{1}{2}} - 0^{\frac{1}{2}}) = 2\)

Therefore, the original expression's limit evaluates to 2. Hence, the correct answer is:

2