JEE Main 2025 — 29 January, Morning Shift — Question 59

$\lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{\mathrm{k}^{3}+6 \mathrm{k}^{2}+11 \mathrm{k}+5}{(\mathrm{k}+3)!}$

$=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{3}+6 k^{2}+11 k+6-1}{(k+3)!}$

$=\lim _{n \rightarrow \infty} \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)-1}{(\mathrm{k}+3)!}$

$=\lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{(\mathrm{k}+3)!}-\frac{1}{(\mathrm{k}+3)!}$

$=\lim _{\mathrm{k}=1} \sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{1}{\mathrm{k}!}-\frac{1}{(\mathrm{k}+3)!}\right)$

$=\lim _{\mathrm{k}=1}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} \ldots+\frac{1}{\mathrm{n}!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!} \ldots-\frac{1}{(\mathrm{n}+3)!}\right)$

$=\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}$