Key Concepts and Formulas

• Limits of Sequences: Understanding how to evaluate the limit of a sequence as $n$ approaches infinity. This often involves identifying the dominant terms in the numerator and denominator.
• Summation Formulas: Knowledge of formulas for the sum of the first $k$ natural numbers ($\sum_{i=1}^k i = \frac{k(k+1)}{2}$) and arithmetic progressions.
• Rationalization of Denominator: A technique used to simplify expressions involving square roots in the denominator by multiplying the numerator and denominator by the conjugate of the denominator.
• Algebraic Manipulation: Proficiency in manipulating algebraic expressions, especially when dealing with large powers of $n$ and simplifying fractions.

Step-by-Step Solution

Let the given limit be denoted by $I$. $I = \mathop {\lim }\limits_{n \to \infty } \frac{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4}}$

Step 1: Analyze and simplify the numerator. The numerator is a series that can be grouped into triplets: $(1+2-3) + (4+5-6) + \dots + ((3n-2)+(3n-1)-3n)$. Each triplet sums to 0: $(3k-2) + (3k-1) - 3k = 3k - 2 + 3k - 1 - 3k = 3k - 3$. This is incorrect. Let's re-examine the pattern. The pattern is $1+2-3$, $4+5-6$, etc. The $k$-th triplet is $(3k-2) + (3k-1) - 3k$. Sum of the $k$-th triplet: $(3k-2) + (3k-1) - 3k = 6k - 3 - 3k = 3k - 3$. This still doesn't seem right as the given example is $1+2-3 = 0$. Let's look at the terms directly. The series is $1+2-3+4+5-6+\dots+(3n-2)+(3n-1)-3n$. We can group the positive terms and the negative terms separately. Sum of positive terms: $1+2+4+5+\dots+(3n-2)+(3n-1)$. Sum of negative terms: $-3-6-9-\dots-3n$.

Let's try grouping the terms differently. The series can be written as: $(1+2) - 3 + (4+5) - 6 + \dots + ((3n-2)+(3n-1)) - 3n$. The sum of the first $n$ triplets is: The $k$-th triplet is $(3k-2) + (3k-1) - 3k$. Sum of the $k$-th triplet: $(3k-2) + (3k-1) - 3k = 6k - 3 - 3k = 3k - 3$. This is incorrect. Let's look at the first few triplets: $1+2-3 = 0$ $4+5-6 = 3$ $7+8-9 = 6$ The sum of the $k$-th triplet is $3(k-1)$. So the numerator is the sum of $n$ such triplets. Numerator = $\sum_{k=1}^{n} (3k-3) = 3 \sum_{k=1}^{n} (k-1) = 3 \sum_{j=0}^{n-1} j = 3 \frac{(n-1)n}{2}$.

Alternatively, let's use the approach from the provided solution to interpret the numerator. Numerator $= (1+2+3+\dots+3n) - 2(3+6+9+\dots+3n)$. Sum of $1+2+\dots+3n = \frac{3n(3n+1)}{2}$. Sum of $3+6+9+\dots+3n = 3(1+2+3+\dots+n) = 3 \frac{n(n+1)}{2}$. So, the numerator is: $\frac{3n(3n+1)}{2} - 2 \left( 3 \frac{n(n+1)}{2} \right) = \frac{3n(3n+1)}{2} - 3n(n+1)$ $= \frac{3n}{2} [(3n+1) - 2(n+1)]$ $= \frac{3n}{2} [3n+1 - 2n - 2]$ $= \frac{3n}{2} [n-1]$.

Step 2: Rationalize the denominator. The denominator is in the form $\sqrt{A} - \sqrt{B}$. We multiply the numerator and denominator by the conjugate $\sqrt{A} + \sqrt{B}$. Denominator = $(\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4}) \times \frac{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}$ $= \frac{(2n^4 + 4n + 3) - (n^4 + 5n + 4)}{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}$ $= \frac{2n^4 + 4n + 3 - n^4 - 5n - 4}{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}$ $= \frac{n^4 - n - 1}{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}$.

Step 3: Substitute the simplified numerator and denominator back into the limit expression. $I = \mathop {\lim }\limits_{n \to \infty } \frac{\frac{3n(n-1)}{2}}{\frac{n^4 - n - 1}{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}}$ $I = \mathop {\lim }\limits_{n \to \infty } \frac{3n(n-1)}{2} \times \frac{\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}}{n^4 - n - 1}$ $I = \mathop {\lim }\limits_{n \to \infty } \frac{3n^2(n-1)[\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}]}{2(n^4 - n - 1)}$

Step 4: Divide the numerator and denominator by the highest power of $n$ in the denominator. The highest power of $n$ in the denominator is $n^4$. Let's extract the highest power of $n$ from each term. From the numerator: $3n^2(n-1) = 3n^3 - 3n^2$. From the square roots in the numerator: $\sqrt{2n^4 + 4n + 3} = \sqrt{n^4(2 + \frac{4}{n^3} + \frac{3}{n^4})} = n^2 \sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}}$ $\sqrt{n^4 + 5n + 4} = \sqrt{n^4(1 + \frac{5}{n^3} + \frac{4}{n^4})} = n^2 \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}$ So, the term with the square roots becomes: $n^2 (\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}})$.

The numerator is: $\frac{3n(n-1)}{2} [\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}]$ $= \frac{3n^2(n-1)}{2} [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]$ $= \frac{3n^3 - 3n^2}{2} [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]$

The denominator is: $n^4 - n - 1$.

Now, let's rewrite the entire limit expression by dividing the numerator and denominator by $n^4$. $I = \mathop {\lim }\limits_{n \to \infty } \frac{\frac{3n^3 - 3n^2}{2} [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{\frac{n^4 - n - 1}{n^4}}$ I = \mathop {\lim }\limits_{n \to \infty } \frac{\frac{3}{n} - \frac{3}{n^2}}{2} [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{\frac{1}{n^4} - \frac{1}{n^3} - \frac{1}{n^4}} This approach seems to be getting complicated. Let's go back to the expression from Step 3 and divide by the appropriate powers.

$I = \mathop {\lim }\limits_{n \to \infty } \frac{3n(n-1)[\sqrt {2{n^4} + 4n + 3} + \sqrt {{n^4} + 5n + 4}]}{2(n^4 - n - 1)}$ Divide numerator and denominator by $n^4$. Numerator: $3n(n-1)[\sqrt {n^4(2 + \frac{4}{n^3} + \frac{3}{n^4})} + \sqrt {n^4(1 + \frac{5}{n^3} + \frac{4}{n^4})}]$ $= 3n(n-1)n^2[\sqrt {2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt {1 + \frac{5}{n^3} + \frac{4}{n^4}}]$ $= 3n^3(n-1)[\sqrt {2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt {1 + \frac{5}{n^3} + \frac{4}{n^4}}]$

Denominator: $2(n^4 - n - 1)$

$I = \mathop {\lim }\limits_{n \to \infty } \frac{3n^3(n-1)[\sqrt {2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt {1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{2(n^4 - n - 1)}$ Divide numerator and denominator by $n^4$. $I = \mathop {\lim }\limits_{n \to \infty } \frac{3n^3(n-1)}{n^4} \times \frac{[\sqrt {2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt {1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{2\frac{(n^4 - n - 1)}{n^4}}$ $I = \mathop {\lim }\limits_{n \to \infty } \frac{3(n-1)}{n} \times \frac{[\sqrt {2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt {1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{2(1 - \frac{1}{n^3} - \frac{1}{n^4})}$ $I = \mathop {\lim }\limits_{n \to \infty } 3\left(1-\frac{1}{n}\right) \times \frac{[\sqrt {2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt {1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{2(1 - \frac{1}{n^3} - \frac{1}{n^4})}$

Step 5: Evaluate the limit by substituting $n \to \infty$. As $n \to \infty$, terms like $\frac{1}{n}$, $\frac{1}{n^3}$, $\frac{4}{n^3}$, $\frac{3}{n^4}$, $\frac{5}{n^3}$, $\frac{4}{n^4}$, $\frac{1}{n^3}$, $\frac{7}{n^4}$ (from the original solution's denominator calculation) approach 0. $I = 3(1-0) \times \frac{[\sqrt{2 + 0 + 0} + \sqrt{1 + 0 + 0}]}{2(1 - 0 - 0)}$ $I = 3 \times \frac{\sqrt{2} + \sqrt{1}}{2}$ $I = 3 \times \frac{\sqrt{2} + 1}{2}$ $I = \frac{3(\sqrt{2} + 1)}{2}$

Let's re-check the denominator calculation from the original solution. The original solution's denominator rationalization step: $\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]$ This seems to have a typo. It should be $(2 n^4+4 n+3) - (n^4+5 n+4)$. So, the denominator after rationalization is: $(2n^4 + 4n + 3) - (n^4 + 5n + 4) = 2n^4 + 4n + 3 - n^4 - 5n - 4 = n^4 - n - 1$. This matches our calculation.

Let's look at the original solution's step 3 and 4 again. $I=\lim _{n \rightarrow \infty} \frac{\frac{3 n(3 n+1)}{2}-6 \frac{n(n+1)}{2}}{\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)}$ $=\lim _{n \rightarrow \infty} \frac{3 n(n-1)\left[\sqrt{2 n^4+4 n+3}+\sqrt{n^4+5 n+4}\right]}{2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]}$ There is a typo in the original solution's denominator calculation: $2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]$. It should be the term from the rationalization: $2 \cdot [(2n^4 + 4n + 3) - (n^4 + 5n + 4)] = 2(n^4 - n - 1)$. Then the original solution proceeds to: $=\lim _{n \rightarrow \infty} \frac{3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)\left[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right]}{2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]}$ Let's trace how they got $3 \cdot 1 \cdot (1-\frac{1}{n})$. The numerator is $\frac{3n(n-1)}{2} [\sqrt{2n^4+\dots} + \sqrt{n^4+\dots}]$. The denominator, after rationalization, is $n^4 - n - 1$. So the expression is: $\frac{\frac{3n(n-1)}{2} [\sqrt{2n^4+\dots} + \sqrt{n^4+\dots}]}{n^4 - n - 1}$ $= \frac{3n(n-1)[\sqrt{2n^4+\dots} + \sqrt{n^4+\dots}]}{2(n^4 - n - 1)}$ Divide numerator and denominator by $n^4$: $= \frac{\frac{3n(n-1)}{n^4}[\sqrt{2n^4+\dots} + \sqrt{n^4+\dots}]}{\frac{2(n^4 - n - 1)}{n^4}}$ $= \frac{3\frac{n^2-n}{n^4}[\sqrt{2n^4+\dots} + \sqrt{n^4+\dots}]}{2(1 - \frac{1}{n^3} - \frac{1}{n^4})}$ $= \frac{3(\frac{1}{n^2} - \frac{1}{n^3})[\sqrt{2n^4+\dots} + \sqrt{n^4+\dots}]}{2(1 - \frac{1}{n^3} - \frac{1}{n^4})}$ This is not matching the original solution. Let's re-examine the original solution's manipulation.

Original solution's step 3: $=\lim _{n \rightarrow \infty} \frac{3 n(n-1)\left[\sqrt{2 n^4+4 n+3}+\sqrt{n^4+5 n+4}\right]}{2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]}$ The denominator calculation in the original solution is $2 \cdot [(2n^4+4n-3) - (n^4+5n+4)] = 2[n^4 - n - 7]$. This is different from our calculation $2(n^4 - n - 1)$. There is a typo in the original solution's denominator terms.

Let's assume the original solution meant to write the correct denominator after rationalization, which is $n^4 - n - 1$. Then the expression is: $I = \mathop {\lim }\limits_{n \to \infty } \frac{\frac{3n(n-1)}{2} [\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}]}{n^4 - n - 1}$ $I = \mathop {\lim }\limits_{n \to \infty } \frac{3n(n-1)[\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}]}{2(n^4 - n - 1)}$ Divide numerator and denominator by $n^4$: $I = \mathop {\lim }\limits_{n \to \infty } \frac{\frac{3n(n-1)}{n^2} \frac{1}{n^2}[\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}]}{\frac{2(n^4 - n - 1)}{n^4}}$ This is not the right way to divide by $n^4$. Let's divide the numerator and denominator by $n^4$. Numerator: $3n(n-1)[\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}]$ Divide by $n^4$: $\frac{3n(n-1)}{n^4} [\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}]$ $= \frac{3(n-1)}{n^3} [n^2\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + n^2\sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]$ $= \frac{3(n-1)}{n} [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]$ $= 3(1-\frac{1}{n}) [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]$

Denominator: $2(n^4 - n - 1)$. Divide by $n^4$: $2(1 - \frac{1}{n^3} - \frac{1}{n^4})$.

So, the limit becomes: $I = \mathop {\lim }\limits_{n \to \infty } \frac{3(1-\frac{1}{n}) [\sqrt{2 + \frac{4}{n^3} + \frac{3}{n^4}} + \sqrt{1 + \frac{5}{n^3} + \frac{4}{n^4}}]}{2(1 - \frac{1}{n^3} - \frac{1}{n^4})}$ As $n \to \infty$: $I = \frac{3(1-0) [\sqrt{2 + 0 + 0} + \sqrt{1 + 0 + 0}]}{2(1 - 0 - 0)}$ $I = \frac{3(1) [\sqrt{2} + \sqrt{1}]}{2(1)}$ $I = \frac{3(\sqrt{2} + 1)}{2}$

The original solution's step 4 has a typo in the denominator: $2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]$. It should be $2\left[1-\frac{1}{n^3}-\frac{1}{n^4}\right]$. And the numerator in original solution's step 4: $3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)$ is also not directly obtained. It should be $3(1-\frac{1}{n})$. The $1$ is likely from $n/n$.

Let's re-examine the original solution's step 4: $=\lim _{n \rightarrow \infty} \frac{3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)\left[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right]}{2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]}$ The numerator term $3 \cdot 1 \cdot (1-\frac{1}{n})$ is likely from $\frac{3n(n-1)}{n^2} \times \frac{n^2}{n^2} \dots$ which is not clear. The most straightforward approach is to divide the numerator and denominator of the entire fraction by $n^4$.

Let's assume the original solution's calculation of the numerator's factor to be correct as $3(1-\frac{1}{n})$. And the square root part: $[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}]$. And the denominator part: $2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]$. If we substitute $n \to \infty$: Numerator $\to 3(1-0)[\sqrt{2+0+0} + \sqrt{1+0+0}] = 3(\sqrt{2}+1)$. Denominator $\to 2[1-0-0] = 2$. The limit would be $\frac{3(\sqrt{2}+1)}{2}$.

My derived result is $\frac{3(\sqrt{2} + 1)}{2}$. The original solution's final answer is $\frac{3(\sqrt{2}+1)}{2}$. The original solution's step 5 is: $=\frac{3(\sqrt{2}+1)}{2}$ This matches my result and the correct answer.

Common Mistakes & Tips

• Numerator Simplification: Carefully group terms in the numerator. Recognize patterns or use summation formulas to simplify it accurately.
• Denominator Rationalization: Ensure the conjugate is correctly identified and applied. Pay close attention to signs during subtraction of terms after rationalization.
• Highest Power of n: When evaluating limits at infinity, identify the highest power of $n$ in both the numerator and denominator and divide accordingly. For expressions involving square roots, remember that $\sqrt{n^k} = n^{k/2}$.

Summary

The problem requires evaluating a limit of a fraction as $n$ approaches infinity. The numerator is a series that can be simplified using summation formulas. The denominator involves square roots, necessitating rationalization. After simplifying both the numerator and the denominator, the limit is evaluated by identifying the dominant terms as $n \to \infty$. The process involves algebraic manipulation, summation formulas, and the technique of rationalizing the denominator.

The final answer is $\boxed{\frac{3(\sqrt{2}+1)}{2}}$.