Question

$\lim _{n \rightarrow \infty} \frac{1 . n+(n-1)(1+2)+(n-2)(1+2+3)+. .1 \cdot \sum_{r=1}^{n} r}{n^{4}}$ is equal to (A) \(1 / 12\) (B) \(1 / 24\) (C) \(1 / 6\) (D) \(1 / 48\)

Short answer

Answer: The value of the limit is 0.

Step-by-step solution

Identify the Sums in the Expression

First, we can rewrite the expression as follows: \(\lim _{n \rightarrow \infty} \frac{n+ (n-1)\sum_{r=1}^{2} r + (n-2)\sum_{r=1}^{3} r+\cdots+1\cdot\sum_{r=1}^{n} r}{n^{4}}\) Now we can focus on finding the sum of the series in the expression.

Compute the Sum for Each Term

The sum of the first 'k' natural numbers can be expressed as: \(\sum_{r=1}^{k} r = \frac{k(k+1)}{2}\) So, our expression becomes: \(\lim _{n \rightarrow \infty} \frac{n+ (n-1)\frac{3}{2} + (n-2)\frac{6}{2}+\cdots+(1)\frac{n(n+1)}{2}}{n^{4}}\)

Factor Out the Common Factor

Now, let's factor out \(\frac{1}{2}\) from each term in the numerator: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(2n+ (n-1)(3) + (n-2)(6)+\cdots+(1)(n(n+1)))}{n^{4}}\)

Simplify the Expression

Now, let's expand the series in the expression: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(2n+ 3n-3 + 6n-12+\cdots+n^{2}+n)}{n^{4}}\) Next, we add the terms: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}((n+1)(n+2)-3n)}{n^{4}}\)

Compute the Limit

Now, divide each term by \(n^{3}\) and find the limit: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(\frac{(n+1)(n+2)}{n^3} - \frac{3n}{n^3})}{n}\) As \(n \rightarrow \infty\), \(\frac{(n+1)(n+2)}{n^3} \rightarrow 1\) and \(\frac{3n}{n^3} \rightarrow 0\): \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(1- 0)}{n}=0\)

Check the Options

Since the value of the limit is 0, none of the options are correct. To double-check this result, you may want to review the problem statement and solution, or perhaps consult your teacher for any errors or mistakes.