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Chapter 9: Problem 716

\(\lim _{\mathrm{n} \rightarrow \infty}\left[\left(1 / \mathrm{n}^{2}\right)+\left(2 / \mathrm{n}^{2}\right)+\left(3 / \mathrm{n}^{2}\right)+\ldots \ldots\left\\{(\mathrm{n}-1) / \mathrm{n}^{2}\right\\}\right]=?\) (a) \((1 / 2)\) (b) 2 (c) (d) \((2 / 3)\)

Short Answer

Expert verified
The short answer is (a) \(\frac{1}{2}\).

Step by step solution

01

Express the sum in Sigma notation

We can express the given sum using the Sigma notation, an indexed summation for a sequence of terms. Note that the terms are of the form \(\frac{k}{n^2}\) where k goes from 1 to (n-1). So, we can write: \[\sum_{k=1}^{n-1} \frac{k}{n^2}\]
02

Factor out n^2

Now, let's factor out the common denominator \(n^2\) from the sum: \[\frac{1}{n^2} \sum_{k=1}^{n-1} k\]
03

Solve the sum (Sum of first n-1 natural numbers)

The sum of the first (n-1) natural numbers has the formula: \[\sum_{k=1}^{n-1} k = \frac{(n-1)(n-2)}{2}\] So, the given expression becomes: \[\frac{1}{n^2} \cdot \frac{(n-1)(n-2)}{2}\]
04

Simplify and find the limit

Now, simplify the expression and find the limit as n approaches infinity: \[\lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{(n-1)(n-2)}{2} = \lim_{n \to \infty} \frac{(n-1)(n-2)}{2 n^{2}}\] Divide numerator and denominator by \(n^2\), it becomes: \[\lim_{n \to \infty} \frac{1 - \frac{1}{n} - \frac{2}{n^2}}{2}\] Now, apply the limit: \[\lim_{n \to \infty} \frac{1 - \frac{1}{n} - \frac{2}{n^2}}{2} = \frac{1 - 0 - 0}{2}\]
05

Final answer

Evaluate the final expression: \[\frac{1}{2}\] So, the answer is (a) \(\frac{1}{2}\).

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Most popular questions from this chapter

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{\left(1+\mathrm{x}^{2}\right)^{(1 / 3)}-(1-2 \mathrm{x})^{(1 / 4)}\right\\} /\left(\mathrm{x}+\mathrm{x}^{2}\right)\right]=?\) (a) \((1 / 3)\) (b) \((1 / 2)\) (c) \((1 / 4)\) (d) \((1 / 12)\)

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{1 /\left[{ }^{3} \sqrt{\left.\left. \left.\left(8 \mathrm{~h}^{3}+\mathrm{h}^{4}\right)\right]\right\\}-(1 / 2 \mathrm{~h})\right]}=?\right.\right.\right.\) (a) \((1 / 2)\) (b) \((1 / 68) \quad\) (c) \(-(1 / 12) \quad\) (d) \(-(1 / 48)\)

If \(\mathrm{f}(\mathrm{x})=\mid \begin{array}{ll}\mathrm{m}+3 \mathrm{nx}, & \mathrm{x}>1 \\ 11, & \mathrm{x}=1 \\ 5 \mathrm{nx}-2 \mathrm{~m}, & \mathrm{x}<1\end{array}\) is continuous at \(\mathrm{x}=1\) then \(\mathrm{m}=\ldots .\) and \(\mathrm{n}=\ldots . .\) ? (a) \(\mathrm{m}=2, \mathrm{n}=-3\) (b) \(\mathrm{m}=-2, \mathrm{n}=3\) (c) \(\mathrm{m}=2, \mathrm{n}=3\) (d) \(m=3, n=3\)

\(\lim _{\mathrm{x} \rightarrow(\pi / 2)}[\\{\sin (\cos \mathrm{x}) \cos \mathrm{x}\\} /\\{\sin \mathrm{x}-\operatorname{cosec} \mathrm{x}\\}]=?\) (a) 0 (b) 1 (c) Limit does not exist (d) \(-1\)

\(\lim _{\theta \rightarrow 0} 2[\\{\sqrt{3} \operatorname{Sin}[(\pi / 6)+\theta]-\operatorname{Cos}[(\pi / 6)+\theta]\\}\) \(/\\{\sqrt{3} \theta(\sqrt{3} \operatorname{Cos} \theta-\operatorname{Sin} \theta)\\}]=?\) (a) \((4 / 3)\) (b) \((\sqrt{3} / 4)\) (c) \((4 / 9)\) (d) \((2 / 3)\)
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