1. We need to evaluate

$$L=\lim_{n\to\infty}\sum_{k=1}^n \frac{k^3+6k^2+11k+5}{(k+3)!}.$$

Since this is an infinite series with factorial in the denominator, we try to rewrite the numerator so that the term becomes telescoping.

2. Let

$$T_k=\frac{k^3+6k^2+11k+5}{(k+3)!}.$$

We look for constants so that

$$k^3+6k^2+11k+5$$

can be expressed in terms of products related to $(k+3)!$.

Notice that

$$(k+3)!=(k+3)(k+2)(k+1)k!.$$

Also,

$$(k+1)(k+2)(k+3)=k^3+6k^2+11k+6.$$

Therefore,

$$k^3+6k^2+11k+5=(k+1)(k+2)(k+3)-1.$$

So,

$$T_k=\frac{(k+1)(k+2)(k+3)-1}{(k+3)!} =\frac{1}{k!}-\frac{1}{(k+3)!}.$$

3. Hence the partial sum becomes

$$S_n=\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right).$$

Split the sums:

$$S_n=\sum_{k=1}^n\frac{1}{k!}-\sum_{k=1}^n\frac{1}{(k+3)!}.$$

In the second sum, put $r=k+3$. Then as $k=1\to n$, $r=4\to n+3$:

$$\sum_{k=1}^n\frac{1}{(k+3)!}=\sum_{r=4}^{n+3}\frac{1}{r!}.$$

Therefore,

$$S_n=\sum_{k=1}^n\frac{1}{k!}-\sum_{r=4}^{n+3}\frac{1}{r!}.$$

Most terms cancel, leaving

$$S_n=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}-\left(\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}\right).$$

4. Now take the limit as $n\to\infty$. Since factorials grow very fast,

$$\frac{1}{(n+1)!},\ \frac{1}{(n+2)!},\ \frac{1}{(n+3)!}\to 0.$$

So,

$$L=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=1+\frac12+\frac16=\frac{5}{3}.$$

5. Check options:

• A: $\frac53$ ✅
• B: $2$ ❌
• C: $\frac43$ ❌
• D: $\frac73$ ❌

Thus the correct answer is A.