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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Sequences and Series

The value of ∑_ k=1 ^ ∞ (-1)^ k+1 ((k(k+1))/(k!)) is

The value of ∑k=1∞(−1)k+1(k(k+1)k!)\sum_{k=1}^{\infty}(-1)^{k+1}\left(\frac{k(k+1)}{k!}\right) is

Options

  1. A2/e
  2. B1/e1 / \mathrm{e}
  3. Ce/2\mathrm{e} / 2
  4. De\sqrt{e}

Correct answer

B. 1/e1 / \mathrm{e}

Step-by-step solution

Evaluate ∑k=1∞(−1)k+1k(k+1)k!\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k+1)}{k!}.
Rewrite as ∑k=1∞(−1)k+1k+1(k−1)!\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k+1}{(k-1)!} and split: ∑k=1∞(−1)k+1k(k−1)!+∑k=1∞(−1)k+11(k−1)!\sum_{k=1}^{\infty} (-1)^{k+1} \frac{k}{(k-1)!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!}.
The first sum, after rewriting k(k−1)!=1(k−2)!+1(k−1)!\frac{k}{(k-1)!} = \frac{1}{(k-2)!} + \frac{1}{(k-1)!} for k≥2k \geq 2, evaluates to 0 using the exponential series ∑j=0∞(−1)jj!=e−1\sum_{j=0}^{\infty} \frac{(-1)^j}{j!} = e^{-1}.
The second sum with substitution j=k−1j = k-1 gives ∑j=0∞(−1)j1j!=e−1=1e\sum_{j=0}^{\infty} (-1)^j \frac{1}{j!} = e^{-1} = \frac{1}{e}.
Therefore, the total is 0+1e=1e0 + \frac{1}{e} = \frac{1}{e}.
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