Sum the series to infinity :
`sqrt(2)- (1)/(sqrt(2))+(1)/(2(sqrt(2)))-(1)/(4sqrt(2))+ ....`

Answer

To find the sum of the series to infinity given by:

\[ \sqrt{2} - \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{1}{4\sqrt{2}} + \ldots \]

we can follow these steps:

### Step 1: Identify the first term (a) and the common ratio (r)
The first term of the series \( a \) is:

\[ a = \sqrt{2} \]

Next, we need to find the common ratio \( r \). The second term of the series is:

\[ \text{Second term} = -\frac{1}{\sqrt{2}} \]

To find \( r \), we can use the formula:

\[ r = \frac{\text{Second term}}{\text{First term}} = \frac{-\frac{1}{\sqrt{2}}}{\sqrt{2}} = -\frac{1}{2} \]

### Step 2: Use the formula for the sum of an infinite geometric series
The formula for the sum \( S \) of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

where \( |r| < 1 \). In our case, we have:

\[ a = \sqrt{2} \quad \text{and} \quad r = -\frac{1}{2} \]

### Step 3: Substitute the values into the formula
Substituting \( a \) and \( r \) into the formula gives:

\[ S = \frac{\sqrt{2}}{1 - \left(-\frac{1}{2}\right)} = \frac{\sqrt{2}}{1 + \frac{1}{2}} = \frac{\sqrt{2}}{\frac{3}{2}} = \sqrt{2} \cdot \frac{2}{3} = \frac{2\sqrt{2}}{3} \]

### Final Answer
Thus, the sum of the series to infinity is:

\[ \boxed{\frac{2\sqrt{2}}{3}} \]