The value of\sum _{j=0}^{\infty }\sum _{i=1}^{\infty }{2}^{j-3i}=\text{\_\_\_\_\_\_\_\_}

The correct answer is 1.

To find the value of the double summation, we analyze the expression as a product of two independent infinite geometric series. In its standard convergent formulation, the terms of the summation can be separated because the indices i and j do not depend on each other.

We can rewrite the double sum as the product of two individual summations:

$$\sum _{j=1}^{\infty }\sum _{i=1}^{\infty }{2}^{-j-i}=\left(\sum _{j=1}^{\infty }{2}^{-j}\right)\times \left(\sum _{i=1}^{\infty }{2}^{-i}\right)$$

Now, we evaluate each of the infinite geometric series independently. Recall the formula for the sum of an infinite geometric series with the first term a and common ratio r (where |r| < 1):

$$S=\frac{a}{1-r}$$

For the first series in j:

$$\sum _{j=1}^{\infty }{2}^{-j}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$

Here, the first term is a = 1/2 and the common ratio is r = 1/2. Evaluating the sum gives:

$$S_1=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$$

Similarly, for the second series in i:

$$\sum _{i=1}^{\infty }{2}^{-i}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$

This series also has the first term a = 1/2 and common ratio r = 1/2, so its sum is:

$$S_2=1$$

Finally, we multiply the two independent sums to find the value of the double summation:

$$\text{Total Value}=S_1\times S_2=1\times 1=1$$