Key Concepts and Formulas

• Limit of a Sum: The limit of a sum can be evaluated by first finding the general term of the sum and then evaluating the limit of the series.
• Telescoping Series: A series where most of the terms cancel out, leaving only the first few and the last few terms. This often occurs when a general term can be expressed as a difference of consecutive terms of a sequence.
• Factorial Properties: Understanding the definition of the factorial function, $n! = n \times (n-1) \times \dots \times 2 \times 1$.
• Algebraic Manipulation: Techniques for simplifying polynomial and rational expressions, particularly involving factorials.

Step-by-Step Solution

Step 1: Analyze the General Term of the Sum We are asked to find the limit of a sum. The general term of the sum is given by $a_k = \frac{k^3+6 k^2+11 k+5}{(k+3)!}$. Our goal is to simplify this expression, ideally into a form that will lead to a telescoping series.

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

Step 2: Manipulate the Numerator to Relate it to the Denominator Observe the denominator $(k+3)!$. We can try to express the numerator in terms of expressions involving $k+1, k+2, k+3$ or related factors. Let's consider the product $(k+1)(k+2)(k+3)$. Expanding this, we get: $(k+1)(k+2)(k+3) = (k^2 + 3k + 2)(k+3)$ $= k^3 + 3k^2 + 2k + 3k^2 + 9k + 6$ $= k^3 + 6k^2 + 11k + 6$

Notice that the numerator $k^3+6 k^2+11 k+5$ is very close to $(k+1)(k+2)(k+3)$. Specifically, $k^3+6 k^2+11 k+5 = (k^3+6 k^2+11 k+6) - 1 = (k+1)(k+2)(k+3) - 1$.

So, we can rewrite the general term as:

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

Step 3: Separate the General Term into Two Parts We can now split the fraction into two parts:

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

Step 4: Simplify the First Part of the General Term Let's simplify the first term. We know that $(k+3)! = (k+3)(k+2)(k+1)k!$.

$\frac{(k+1)(k+2)(k+3)}{(k+3)!} = \frac{(k+1)(k+2)(k+3)}{(k+3)(k+2)(k+1)k!}$ After cancellation, this simplifies to:

$\frac{1}{k!}$

Step 5: Rewrite the General Term in a Telescoping Form Substituting the simplified first term back into the expression for $a_k$:

$a_k = \frac{1}{k!} - \frac{1}{(k+3)!}$ This form is crucial because it suggests a telescoping series.

Step 6: Evaluate the Sum of the Series We need to find the limit of $\sum_{k=1}^n a_k$ as $n \rightarrow \infty$. Let's write out the terms of the sum:

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

Let's expand the sum for a few terms to see the cancellation: For $k=1$: $\frac{1}{1!} - \frac{1}{4!}$ For $k=2$: $\frac{1}{2!} - \frac{1}{5!}$ For $k=3$: $\frac{1}{3!} - \frac{1}{6!}$ For $k=4$: $\frac{1}{4!} - \frac{1}{7!}$ For $k=5$: $\frac{1}{5!} - \frac{1}{8!}$ ... For $k=n-2$: $\frac{1}{(n-2)!} - \frac{1}{(n+1)!}$ For $k=n-1$: $\frac{1}{(n-1)!} - \frac{1}{(n+2)!}$ For $k=n$: $\frac{1}{n!} - \frac{1}{(n+3)!}$

When we sum these terms, the $-\frac{1}{4!}$ from the first term cancels with the $+\frac{1}{4!}$ from the fourth term. Similarly, $-\frac{1}{5!}$ cancels with $+\frac{1}{5!}$, and so on. The terms that do not cancel are the initial positive terms and the final negative terms.

The sum becomes: $\sum_{k=1}^n \left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) = \left(\frac{1}{1!} - \frac{1}{4!}\right) + \left(\frac{1}{2!} - \frac{1}{5!}\right) + \left(\frac{1}{3!} - \frac{1}{6!}\right) + \left(\frac{1}{4!} - \frac{1}{7!}\right) + \dots + \left(\frac{1}{n!} - \frac{1}{(n+3)!}\right)$ The terms $-\frac{1}{4!}, -\frac{1}{5!}, -\frac{1}{6!}$ will be cancelled by the terms $\frac{1}{4!}, \frac{1}{5!}, \frac{1}{6!}$ from later in the series. The terms that remain are the first three positive terms and the last three negative terms.

$\sum_{k=1}^n \left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} - \frac{1}{(n+1)!} - \frac{1}{(n+2)!} - \frac{1}{(n+3)!}$

Step 7: Evaluate the Limit as $n \rightarrow \infty$ Now we take the limit of the sum as $n \rightarrow \infty$.

$\lim_{n \rightarrow \infty} \sum_{k=1}^n \left(\frac{1}{k!} - \frac{1}{(k+3)!}\right) = \lim_{n \rightarrow \infty} \left(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} - \frac{1}{(n+1)!} - \frac{1}{(n+2)!} - \frac{1}{(n+3)!}\right)$

As $n \rightarrow \infty$, the terms $\frac{1}{(n+1)!}$, $\frac{1}{(n+2)!}$, and $\frac{1}{(n+3)!}$ all approach 0 because the factorial in the denominator grows infinitely large.

$\lim_{n \rightarrow \infty} \frac{1}{(n+1)!} = 0$ $\lim_{n \rightarrow \infty} \frac{1}{(n+2)!} = 0$ $\lim_{n \rightarrow \infty} \frac{1}{(n+3)!} = 0$

Therefore, the limit of the sum is:

$\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} - 0 - 0 - 0 = \frac{1}{1} + \frac{1}{2} + \frac{1}{6}$

Step 8: Calculate the Final Value Now, we compute the sum of these fractions:

$1 + \frac{1}{2} + \frac{1}{6} = \frac{6}{6} + \frac{3}{6} + \frac{1}{6} = \frac{6+3+1}{6} = \frac{10}{6}$ Simplifying the fraction:

$\frac{10}{6} = \frac{5}{3}$

Common Mistakes & Tips

• Algebraic Errors: Incorrectly expanding the polynomial in the numerator or making mistakes when simplifying the factorial terms can lead to wrong answers. Always double-check your algebraic manipulations.
• Identifying the Telescoping Pattern: The key to solving this problem efficiently is recognizing that the general term can be written as a difference of two terms, allowing for cancellation. If you don't see this pattern, the problem becomes much harder.
• Handling the First Few Terms: In a telescoping sum, it's important to correctly identify which terms remain. For a difference of the form $f(k) - f(k+m)$, the first $m$ positive terms and the last $m$ negative terms usually remain. In this case, $m=3$, so the first 3 positive terms and the last 3 negative terms are relevant for the infinite sum.

Summary

The problem requires evaluating the limit of a series. The core strategy involves simplifying the general term of the series by manipulating the numerator to match factors in the denominator. By rewriting the numerator as $(k+1)(k+2)(k+3) - 1$, we can split the general term into $\frac{1}{k!} - \frac{1}{(k+3)!}$. This form reveals a telescoping series. Upon summing the series and taking the limit as $n \rightarrow \infty$, the terms involving $n$ vanish, leaving the sum of the first few non-cancelling terms: $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}$. Calculating this sum yields $\frac{5}{3}$.

The final answer is \boxed{5/3}.