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: Fix $i$ and sum over $j$ first.
For a fixed value of $i$, the inner sum is $\sum_{j=1}^{\infty} 2^{-i} 3^{-j} = 2^{-i} \sum_{j=1}^{\infty} 3^{-j}$, since $2^{-i}$ does not depend on $j$ and can be pulled out.

Step 2: Evaluate the inner geometric series.
$\sum_{j=1}^{\infty} 3^{-j} = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$ is geometric with first term $1/3$ and ratio $1/3$, so it sums to $\frac{1/3}{1-1/3} = \frac{1}{2}$. So the inner sum for fixed $i$ becomes $2^{-i} \cdot \frac{1}{2}$.

Step 3: Sum the result over all $i$ from 0 to infinity.
\[ \sum_{i=0}^{\infty} 2^{-i} \cdot \frac{1}{2} = \frac{1}{2}\sum_{i=0}^{\infty} 2^{-i} \]
The remaining series $\sum_{i=0}^{\infty} 2^{-i} = 1 + \frac{1}{2} + \frac{1}{4} + \cdots$ is geometric with first term 1 and ratio $1/2$, giving $\frac{1}{1-1/2} = 2$.

Step 4: Combine.
\[ \frac{1}{2} \times 2 = 1 \]

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