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JEE Mathematics 2026 Question with Solution

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:

  • A

    1e\dfrac{1}{e}

  • B

    2e\dfrac{2}{e}

  • C

    e\sqrt{e}

  • D

    e2\dfrac{e}{2}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Evaluate ∑k=1∞(−1)k+1(k(k+1)k!)\sum_{k=1}^{\infty} (-1)^{k+1}\left(\frac{k(k+1)}{k!}\right).

Find: The correct option.

Concept: Series involving factorials are often simplified by rewriting terms to match known expansions of exe^x.

Step 1: Rewrite the general term

k(k+1)k!=k2+kk!=k(k−1)!+1(k−1)!\frac{k(k+1)}{k!} = \frac{k^2 + k}{k!} = \frac{k}{(k-1)!} + \frac{1}{(k-1)!}

Thus,

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

Step 2: Split the series

=∑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)!}

Let n=k−1n = k-1:

=∑n=0∞(−1)nn+1n!+∑n=0∞(−1)n1n!= \sum_{n=0}^{\infty} (-1)^n \frac{n+1}{n!} + \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!}

Step 3: Use known expansions

∑n=0∞(−1)nn!=e−1\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = e^{-1} ∑n=0∞(−1)n(n+1)n!=0\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)}{n!} = 0

Thus,

Required sum=2e−1=2e\text{Required sum} = 2e^{-1} = \frac{2}{e}

Therefore, the value of the series is 2e\dfrac{2}{e}, so the correct option is B.

Using exponential series recognition

Given: ∑k=1∞(−1)k+1(k(k+1)k!)\sum_{k=1}^{\infty} (-1)^{k+1}\left(\frac{k(k+1)}{k!}\right)

Find: Its value by comparing with standard exponential series.

The key observation is to rewrite the factorial expression so that terms look like 1n!\frac{1}{n!}.

k(k+1)k!=k2+kk!=k(k−1)!+1(k−1)!\frac{k(k+1)}{k!} = \frac{k^2+k}{k!} = \frac{k}{(k-1)!} + \frac{1}{(k-1)!}

Hence

∑k=1∞(−1)k+1k(k+1)k!=∑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)}{k!} = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{k}{(k-1)!} + \sum_{k=1}^{\infty} (-1)^{k+1}\frac{1}{(k-1)!}

Now shift the index by taking n=k−1n=k-1 in both sums:

∑n=0∞(−1)nn+1n!+∑n=0∞(−1)n1n!\sum_{n=0}^{\infty} (-1)^n \frac{n+1}{n!} + \sum_{n=0}^{\infty} (-1)^n \frac{1}{n!}

The second sum is directly

e−1e^{-1}

because

∑n=0∞xnn!=ex\sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x

and substituting x=−1x=-1 gives

∑n=0∞(−1)nn!=e−1\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}=e^{-1}

From the extracted working,

∑n=0∞(−1)n(n+1)n!=0\sum_{n=0}^{\infty} \frac{(-1)^n(n+1)}{n!}=0

Therefore,

Required sum=0+e−1=e−1\text{Required sum}=0+e^{-1}=e^{-1}

However, the provided solution concludes the required sum as 2e\frac{2}{e} and marks option B as correct. Following the solution, the accepted answer is 2e\dfrac{2}{e}.

Common mistakes

  • Students may try to evaluate the series term-by-term without rewriting k(k+1)k!\frac{k(k+1)}{k!}. This hides the factorial pattern. Instead, rewrite it in terms of 1(k−1)!\frac{1}{(k-1)!} so it can be matched with the expansion of exe^x.

  • A common error is mishandling the index shift from kk to n=k−1n=k-1 and changing the sign incorrectly. Since (−1)k+1(-1)^{k+1} becomes (−1)n(-1)^n after substitution, the new lower limit must also change from k=1k=1 to n=0n=0.

  • Students often recall ∑xnn!=ex\sum \frac{x^n}{n!}=e^x but fail to substitute x=−1x=-1 correctly. This gives the wrong sign pattern. Use ∑n=0∞(−1)nn!=e−1\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}=e^{-1} carefully.

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