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Chapter 1: Problem 58

$\lim _{\mathrm{n} \rightarrow \infty}\left\\{\frac{7}{10}+\frac{29}{10^{2}}+\frac{133}{10^{3}}+\ldots . .+\frac{5^{\mathrm{n}}+2^{\mathrm{n}}}{10^{\mathrm{n}}}\right\\}$ equals (A) \(3 / 4\) (B) 2 (C) \(5 / 4\) (D) \(1 / 2\)

Short Answer

Expert verified
From the step-by-step solution, the limit of the given infinite series is equal to the sum of two separate geometric series. After calculating the sum of these two series, we find that the value of the limit is 3/4. So, the answer to this short answer question is: "3/4"

Step by step solution

01

Separate the terms into two geometric series

To do this, rewrite the given sequence as the sum of two separate series: one containing the terms \(\frac{5^n}{10^n}\) and the other containing the terms \(\frac{2^n}{10^n}\). So, the given series becomes: \(\lim_{n \rightarrow \infty} \left( \sum_{n=1}^{\infty} \left( \frac{5^n}{10^n} + \frac{2^n}{10^n}\right)\right)\).
02

Calculate the sum of the first geometric series

The first geometric series is \(\sum_{n=1}^{\infty} \frac{5^n}{10^n}\). To find its sum, find the first term (\(a_1\)) and the common ratio (\(r_1\)). For this series, \(a_1 = \frac{5^1}{10^1} = \frac{1}{2}\) and \(r_1 = \frac{5}{10} = \frac{1}{2}\). Now, use the formula for the sum of an infinite geometric series: \(S_1 = \frac{a_1}{1 - r_1}\). Substitute the values we've found: \(S_1 = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{1}{2}\).
03

Calculate the sum of the second geometric series

The second geometric series is \(\sum_{n=1}^{\infty} \frac{2^n}{10^n}\). Again, find the first term (\(a_2\)) and the common ratio (\(r_2\)). For this series, \(a_2 = \frac{2^1}{10^1} = \frac{1}{5}\) and \(r_2 = \frac{2}{10} = \frac{1}{5}\). Now, use the formula for the sum of an infinite geometric series: \(S_2 = \frac{a_2}{1 - r_2}\). Substitute the values we've found: \(S_2 = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{1}{4}\).
04

Add the sums of the two geometric series

Now that we have the sums of the two geometric series, we can add them together to find the value of the given limit. \(\lim_{n \rightarrow \infty} \left(\sum_{n=1}^{\infty} \left( \frac{5^n}{10^n} + \frac{2^n}{10^n} \right)\right) = S_1 + S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\). So the correct answer is (A) \(3 / 4\).

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Most popular questions from this chapter

Suppose that a and \(\mathrm{b}\) are real positive numbers then the value of \(\lim _{t \rightarrow 0}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1 / t}\) has the value equals to (A) $\frac{\mathrm{a} \ln \mathrm{b}-\mathrm{b} \ln \mathrm{a}}{\mathrm{b}-\mathrm{a}}$ (B) $\frac{\mathrm{b} \ln \mathrm{b}-\mathrm{a} \ln \mathrm{a}}{\mathrm{b}-\mathrm{a}}$ (C) \(\mathrm{b} \ln \mathrm{b}-\mathrm{a} \ln \mathrm{a}\) (D) \(\left(\frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}\)

Assertion (A): An equilateral triangle is filled with n, rows of congruent circles. The limit of the ratio of area of circle to the area of triangle as \(\mathrm{n} \rightarrow \infty\) is \(\frac{\sqrt{3} \pi}{6}\) Reason (R): Let the triangle have side length 1 and radius of circles be \(r\). Then \(2(n-1) r+2 r \sqrt{3}\) \(=1 .\) There are \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) circles, the area ratio \(=\frac{\pi}{2 \sqrt{3}} \frac{n(n-1)}{(n+(\sqrt{3}-1))^{2}}\) which approaches \(\frac{\sqrt{3} \pi}{6}\) as \(\mathrm{n} \rightarrow \infty\).

For which of the following functions, Approx \(\mathrm{f}(\mathrm{x})\) exists : (A) \(\underset{x \rightarrow 1}{\text { Approx }} \frac{x^{2}-1}{|x-1|}\) (B) Approx \(\frac{2\\{x\\}-4}{[x]-3}\) (C) $\underset{x \rightarrow 0}{\operatorname{Approx}} \frac{1}{2-2^{\frac{1}{x}}}$ (D) None of these

Let $\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\left(\frac{3}{\pi} \tan ^{-1} 2 \mathrm{x}\right)^{2 \mathrm{n}}+5}$. Then the set of values of \(x\) for which \(f(x)=0\), is (A) \(|2 x|>\sqrt{3}\) (B) \(|(2 x)|<\sqrt{3}\) (C) \(|2 x| \geq \sqrt{3}\) (D) \(|2 x| \leq \sqrt{3}\)

$\lim _{x \rightarrow-\infty}\left\\{x+\sqrt{x^{2}+3 x \cos \frac{1}{|x|}}\right\\}$ is equal to (A) \(3 / 2\) (B) \(-3 / 2\) (C) \(-1\) (D) none of these
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