The value of lim_{n→∞} (∑_{k=1}^{n} (k^3 + 6k^2 + 11k + 5) / (k+3)!) is equal to:
(1) 4/3
(2) 2
(3)

Q: Why did we choose to manipulate the numerator instead of using partial fractions?

Partial fractions are great for rational functions, but when factorials are involved, they often lead to dead ends. Telescoping is the 'golden key' for factorial series because it allows us to cancel out the vast majority of terms, leaving only the boundaries.

Q: How do I know which terms will cancel in a telescoping sum?

The best way is to write out the first three terms and the last three terms of the summation. By physically writing them down, you will see the 'overlap' where the positive part of one term cancels the negative part of the next. Never trust your mental algebra alone; write it out!

Q: Why does the limit of the tail terms go to zero?

As $n$ approaches infinity, the denominator $(n+3)!$ grows at an incredibly rapid rate—much faster than any polynomial or exponential function. Since the numerator is a constant (1), the fraction $\frac{1}{(n+3)!}$ becomes infinitesimally small, effectively vanishing into zero.

Question

The value of $\lim_{n\rightarrow\infty}(\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{(k+3)!})$ is equal to:

Options:

(A) $\frac{4}{3}$

(B) 2

(C) $\frac{7}{3}$

(D) $\frac{5}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of JEE Advanced mathematics. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of polynomials and factorials.

But beneath that complexity lies a beautiful, rhythmic structure waiting to be revealed. Let us begin by examining our general term:

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

When you encounter a polynomial divided by a factorial, your mathematical intuition should immediately trigger a specific alarm: **Telescoping Series**. This is the art of breaking a complex term into a difference of two simpler terms, such that the intermediate values cancel out like falling dominoes.

### The Algebraic Surgery

Our first mission is to perform some algebraic surgery on that numerator. We want to see if we can relate $k^3 + 6k^2 + 11k + 5$ to the factorial in the denominator.

Consider the product of three consecutive integers: $(k+1)(k+2)(k+3)$. If you expand this, you get $k^3 + 6k^2 + 11k + 6$.

Look closely at how similar this is to our numerator! The difference is merely a constant. We can rewrite our numerator as:

$$(k+1)(k+2)(k+3) - 1$$

Now, substitute this back into our general term. We get:

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

By splitting this into two fractions, we obtain:

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

### The Magic of Cancellation

Here is where the magic happens. Recall that $(k+3)! = (k+3)(k+2)(k+1)k!$.

When we divide the first fraction, the terms $(k+1)(k+2)(k+3)$ cancel out perfectly, leaving us with $\frac{1}{k!}$. Thus, our general term simplifies to:

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

This is the **telescoping form** we were hunting for. Now, we set up the sum:

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

When you write out the terms of this sum, you will see the positive part of one term cancelling the negative part of another. For instance, the $\frac{1}{4!}$ from the first part will cancel with the $-\frac{1}{4!}$ from the second part.

This continues until only the first three terms of the first part and the last three terms of the second part remain.

### The Final Push

As $n$ approaches infinity, the terms involving $(n+1)!$, $(n+2)!$, and $(n+3)!$ in the denominator grow infinitely large, causing the fractions to vanish to zero.

We are left with the sum of the first three terms:

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

Calculating this, we get:

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

You have just mastered the art of the telescoping series. The final result is **$\frac{5}{3}$**. Keep this pattern in your toolkit; it is a powerful weapon for your JEE arsenal.

Animated Solution for Mathematics - Sequence and Series: The value of $lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k ^{3} + 6 k ^{2} + 11 k + 5}{( k + 3 )!})$ is equal to:

Select Answer:

Visualized Solution

Analyze the General Term $T_{k}$

Factoring the Numerator

Adjusting the Constant Term

Splitting the General Term

Simplifying the First Fraction

The Telescoping Form of $T_{k}$

Setting up the Summation $S_{n}$

Expanding the Series

Cancellation of Terms

The Remaining Terms of $S_{n}$

Applying the Limit as $n \to \infty$

Final Arithmetic Calculation

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Surgery

The Magic of Cancellation

The Final Push

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limn→∞​(∑k=1n​(k+3)!k3+6k2+11k+5​) is equal to:

Select Answer:

Visualized Solution

Analyze the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk​

Factoring the Numerator

Adjusting the Constant Term

Splitting the General Term

Simplifying the First Fraction

The Telescoping Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk​

Setting up the Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Expanding the Series

Cancellation of Terms

The Remaining Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Applying the Limit as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n→∞