GATE 2026 DA Q.35: solution and answer

GATE 2026 · DAQ.35 · 1 mark · NAT

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

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Answer and solution

ANSWER

The value of the double sum is 1.

Official answer: 1Source: Official answer key, GATE 2026 DA (IIT Guwahati)

Every term is positive and the summand $2^{- i} 3^{- j}$ is a function of $i$ times a function of $j$ , with independent index ranges. So the double series splits into a product of two single series: $i = 0 \sum \infty j = 1 \sum \infty 2^{- i} 3^{- j} = (i = 0 \sum \infty 2^{- i}) (j = 1 \sum \infty 3^{- j}) .$
The first factor is a geometric series with first term $1$ and ratio $\frac{1}{2}$ . Since $∣ r ∣ < 1$ it converges: $i = 0 \sum \infty (\frac{1}{2})^{i} = \frac{1}{1 - \frac{1}{2}} = 2.$
The second factor is a geometric series that starts at $j = 1$ , with first term $\frac{1}{3}$ and ratio $\frac{1}{3}$ : $j = 1 \sum \infty (\frac{1}{3})^{j} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1/3}{2/3} = \frac{1}{2} .$
Multiply the two sums: $2 \times \frac{1}{2} = 1$ .

KEY IDEA

A double sum of positive terms whose summand factors into a function of $i$ times a function of $j$ equals the product of the two single sums. Each factor here is a geometric series, $\sum_{k \geq k_{0}} r^{k} = r^{k_{0}} / (1 - r)$ for $∣ r ∣ < 1$ .

WATCH OUT FOR

Using the formula $1/ (1 - r)$ for the $j$ series, which starts at $j = 1$ , not $j = 0$ . That gives $3/2$ in place of $1/2$ and a final answer of $3$ . The first term of the $j$ series is $\frac{1}{3}$ , not $1$ .

From the GATE 2026 DA paper. Every question of this paper · Download the full paper · All 2 Taylor series questions, with notes