The sum of the infinite series
`"sin"^(-1)((1)/(sqrt2))+"sin"^(-1)((sqrt2-1)/(sqrt6))+"sin"^(-1)((sqrt3-sqrt2)/(sqrt12))+…+…+"sin"^(-1)((sqrtn-sqrt((n-1)))/(sqrt(n(n+1))))+…` is

Options

A. `pi/8`
B. `pi/4`
C. `pi/2`
D. `pi`

Answer

To find the sum of the infinite series

\[ \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + \sin^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right) + \sin^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right) + \ldots + \sin^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right) + \ldots \]

we will follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:

\[ T_n = \sin^{-1}\left(\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\right) \]

### Step 2: Simplify the General Term
We can simplify the term inside the sine inverse function:

\[ T_n = \sin^{-1}\left(\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}\right) \]

Using the identity for sine inverse, we can express this as:

\[ T_n = \sin^{-1}\left(\frac{1}{\sqrt{n}}\right) - \sin^{-1}\left(\frac{1}{\sqrt{n+1}}\right) \]

### Step 3: Write the Series in Summation Form
The series can now be rewritten as:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left(\sin^{-1}\left(\frac{1}{\sqrt{n}}\right) - \sin^{-1}\left(\frac{1}{\sqrt{n+1}}\right)\right) \]

### Step 4: Recognize the Telescoping Nature
This series is telescoping, meaning that many terms will cancel out:

\[ S = \left(\sin^{-1}\left(1\right) - \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) + \left(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) - \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\right) + \left(\sin^{-1}\left(\frac{1}{\sqrt{3}}\right) - \sin^{-1}\left(\frac{1}{\sqrt{4}}\right)\right) + \ldots \]

### Step 5: Evaluate the Limit
As \( n \to \infty \), \( \sin^{-1}\left(\frac{1}{\sqrt{n+1}}\right) \to 0 \). Therefore, the series simplifies to:

\[ S = \sin^{-1}(1) - 0 = \frac{\pi}{2} \]

### Conclusion
Thus, the sum of the infinite series is:

\[ \boxed{\frac{\pi}{2}} \]