Question

The value of \(\displaystyle\sum_{i=0}^{\infty}\sum_{j=1}^{\infty} 2^{-i}\,3^{-j}\) is ________. (Answer in integer)

Hint:

Split the double sum into a product of two independent geometric series, one in i and one in j.

Answer 1

Correct Answer: 1

Step 1: Split the double sum into a product.
The summand \(2^{-i} 3^{-j}\) is a product of a term that depends only on \(i\) and a term that depends only on \(j\). Since \(i\) and \(j\) range over independent, separate index sets, the double sum splits into a product of two single sums:
\[ \sum_{i=0}^{\infty}\sum_{j=1}^{\infty} 2^{-i}\,3^{-j} = \left(\sum_{i=0}^{\infty} 2^{-i}\right)\left(\sum_{j=1}^{\infty} 3^{-j}\right) \]

Step 2: Evaluate the first geometric series.
\(\sum_{i=0}^{\infty} 2^{-i}\) is a geometric series starting at \(i=0\) with first term 1 and common ratio \(1/2\). Its sum is
\[ \sum_{i=0}^{\infty} 2^{-i} = \frac{1}{1 - 1/2} = 2 \]

Step 3: Evaluate the second geometric series.
\(\sum_{j=1}^{\infty} 3^{-j}\) is a geometric series starting at \(j=1\) with first term \(1/3\) and common ratio \(1/3\). Its sum is
\[ \sum_{j=1}^{\infty} 3^{-j} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2} \]

Step 4: Multiply the two results.
\[ 2 \times \frac{1}{2} = 1 \]

Final Answer:
\[ \boxed{1} \]