JEE Main 2026 — Mathematics Algebra

Evaluate $\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k+1)}{k!}$k=1∑∞​(−1)k+1k!k(k+1)​.

Rewrite as $\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k+1}{(k-1)!}$k=1∑∞​(−1)k+1(k−1)!k+1​ and split: $\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k}{(k-1)!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!}$k=1∑∞​(−1)k+1(k−1)!k​+k=1∑∞​(−1)k+1(k−1)!1​.

The first sum, after rewriting $\frac{k}{(k-1)!} = \frac{1}{(k-2)!} + \frac{1}{(k-1)!}$(k−1)!k​=(k−2)!1​+(k−1)!1​ for $k \geq 2$k≥2, evaluates to 0 using the exponential series $\displaystyle \sum_{j=0}^{\infty} \frac{(-1)^j}{j!} = e^{-1}$j=0∑∞​j!(−1)j​=e−1.

The second sum with substitution $j = k-1$j=k−1 gives $\displaystyle \sum_{j=0}^{\infty} (-1)^j \frac{1}{j!} = e^{-1} = \frac{1}{e}$j=0∑∞​(−1)jj!1​=e−1=e1​.

Therefore, the total is $0 + \frac{1}{e} = \frac{1}{e}$0+e1​=e1​.