JEE Main 2026 — 28 January, Morning Shift — Question 2

We need to evaluate $S = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k+1)}{k!}$.

First, rewrite $k(k+1) = k^2 + k$.

So $S = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k^2 + k}{k!} = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k^2}{k!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k}{k!}$.

Now, $\frac{k}{k!} = \frac{1}{(k-1)!}$ and $\frac{k^2}{k!} = \frac{k}{(k-1)!} = \frac{(k-1)+1}{(k-1)!} = \frac{1}{(k-2)!} + \frac{1}{(k-1)!}$ for $k \ge 2$.

Thus, $S = \sum_{k=2}^{\infty} (-1)^{k+1} \frac{1}{(k-2)!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!}$.

Shift indices: let $j = k-2$ in first sum, $j = k-1$ in others.

First sum: $\sum_{j=0}^{\infty} (-1)^{j+3} \frac{1}{j!} = -\sum_{j=0}^{\infty} \frac{(-1)^j}{j!} = -e^{-1}$. Second sum: $\sum_{j=0}^{\infty} (-1)^{j+2} \frac{1}{j!} = \sum_{j=0}^{\infty} \frac{(-1)^j}{j!} = e^{-1}$. Third sum: same as second = $e^{-1}$.

Therefore, $S = -e^{-1} + e^{-1} + e^{-1} = e^{-1} = \frac{1}{e}$.