Chapter 1: Q5MP (page 1)

Test for convergence: $\sum_{n = 2}^{\infty} \frac{{(n - 1)}^{2}}{1 + n^{2}}$

Short Answer

Expert verified

The series $\sum_{n = 2}^{\infty} \frac{{(n - 1)}^{2}}{1 + n^{2}}$ diverges.

Step by step solution

Concept used to prove that the series diverges

Conditions of convergence for the series $\sum_{n = 1}^{\infty} a_{n}$ are:

$l i m_{n \to \infty} a_{n} = 0$

Calculation used to show that the series ∑n=2∞(n-1)21+n2 diverges

$n^{th}$ term of the series is, $a_{n} = \frac{{(n - 1)}^{2}}{1 + n^{2}}$

Apply limit in $n^{th}$ terms of the series as follows:

$l i m_{n \to \infty} a_{n} = l i m_{n \to \infty} \frac{{(n - 1)}^{2}}{1 + n^{2}} l i m_{n \to \infty} a_{n} = l i m_{n \to \infty} \frac{n^{2} {(1 - \frac{1}{n})}^{2}}{n^{2} (\frac{1}{n^{2}} + 1)} l i m_{n \to \infty} a_{n} = l i m_{n \to \infty} \frac{{(1 - \frac{1}{w})}^{2}}{(\frac{1}{a^{2}} + 1)} l i m_{n \to \infty} a_{n} = \frac{1}{1} l i m_{n \to \infty} a_{n} = 1$

Thus, this does not prove the second condition of the convergence as $l i m_{n \to \infty} a_{n} \neq 0$ .

Hence, the series diverges.

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Test for convergence: $\sum_{n = 2}^{\infty} \frac{{(n - 1)}^{2}}{1 + n^{2}}$

Short Answer

Step by step solution

Concept used to prove that the series diverges

Calculation used to show that the series ∑n=2∞(n-1)21+n2 diverges

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Test for convergence:∑n=2∞(n-1)21+n2

Short Answer

Step by step solution

Concept used to prove that the series diverges

Calculation used to show that the series ∑n=2∞(n-1)21+n2 diverges

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Scientific Method Physics

Nuclear Physics

Thermodynamics

Circular Motion and Gravitation

Oscillations

Electric Charge Field and Potential

Study anywhere. Anytime. Across all devices.

Test for convergence: $\sum_{n = 2}^{\infty} \frac{{(n - 1)}^{2}}{1 + n^{2}}$