The sum of the infinte series `sin^(-1)(1/sqrt(2))+sin^(-1)((sqrt(2)-1)/(sqrt(6)))+....sin^(-1)((sqrt(n)-sqrt(n-1))/(sqrt(n(n+1))))`

Answer

To solve the infinite series

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

we can denote the general term of the series as:

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

### Step 1: Rewrite the term

We can rewrite \(T_n\) as:

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

This can be expressed as:

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

### Step 2: Use the sine inverse subtraction formula

Using the formula for the difference of inverse sine:

\[ \sin^{-1}(x) - \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} - y\sqrt{1-x^2}\right) \]

We can express \(T_n\) as:

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

### Step 3: Sum the series

Now we can sum the series:

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

This is a telescoping series, where most terms will cancel out:

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

### Step 4: Evaluate the limit

As \(n\) approaches infinity, \(\sin^{-1}\left(\frac{1}{\sqrt{n+1}}\right)\) approaches \(\sin^{-1}(0) = 0\).

Thus, we have:

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

### Final Answer

The sum of the infinite series is:

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