GATE 2026 DA Q.45: solution and answer

GATE 2026 · DAQ.45 · 2 marks · MCQ

Let $L = n \to \infty lim k = 0 \sum n \frac{e ^{- n} n ^{k}}{k !}$ Which of the following is the value of $L$ ?

(A)
0.5
(B)
1.0
(C)
0
(D)
$e^{- 1}$

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Answer and solution

ANSWER

$L = \frac{1}{2} = 0.5$ , option (A).

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

The term $\frac{e ^{- n} n ^{k}}{k !}$ is the probability mass function of a $Poisson (n)$ random variable at $k$ . So the sum equals $P (X_{n} \leq n)$ , where $X_{n} \sim Poisson (n)$ .
A $Poisson (n)$ variable has the same distribution as the sum of $n$ independent $Poisson (1)$ variables $Y_{1}, \dots, Y_{n}$ , because Poisson variables add by adding their means. Each $Y_{i}$ has mean $1$ and variance $1$ , so $E [X_{n}] = n$ and $Var (X_{n}) = n$ .
Standardise: $P (X_{n} \leq n) = P (\frac{X _{n} - n}{n} \leq 0)$ .
By the central limit theorem, $\frac{X _{n} - n}{n} \to Z \sim N (0, 1)$ in distribution as $n \to \infty$ .
Convergence in distribution gives convergence of the cdf at every continuity point of the limit cdf. The normal cdf is continuous everywhere, in particular at $0$ , so $L = Φ (0) = \frac{1}{2}$ .
Hence $L = 0.5$ , which is option (A).

KEY IDEA

The sum is the Poisson( $n$ ) cdf evaluated at its own mean $n$ . A Poisson( $n$ ) variable is a sum of $n$ independent Poisson(1) variables, so the central limit theorem makes it approximately normal, and the normal cdf at its mean is $\frac{1}{2}$ .

WATCH OUT FOR

Reading the sum as the whole Poisson series and answering $1$ (option B). That would need the sum to run to infinity. It stops at $k = n$ , which is the mean, so only about half the probability mass is included. Another trap is answering $e^{- 1}$ by looking only at the $e^{- n}$ factor.

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