Question

Math

Posted 3 months ago

```
6. Which of the following statements about the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}}$ is true?
(A) The series converges absolutely.
(B) The series converges conditionally.
(C) The series converges but neither conditionally nor absolutely.
(D) The series diverges.
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 1

Consider the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}}$. We need to determine its convergence properties

step 2

To check for absolute convergence, we examine the series $\sum_{n=1}^{\infty} \left| \frac{(-1)^{n}}{1+\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}}$

step 3

Note that $\frac{1}{1+\sqrt{n}}$ is a positive, decreasing function. We compare it to the integral $\int_{1}^{\infty} \frac{1}{1+\sqrt{x}} \, dx$

step 4

Evaluate the integral $\int_{1}^{\infty} \frac{1}{1+\sqrt{x}} \, dx$. Let $u = \sqrt{x}$, then $du = \frac{1}{2\sqrt{x}} \, dx$ and $dx = 2u \, du$

step 5

The integral becomes $\int_{1}^{\infty} \frac{2u \, du}{1+u}$. This integral diverges because the integrand behaves like $\frac{2u}{u} = 2$ for large $u$

step 6

Since the integral diverges, the series $\sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}}$ does not converge absolutely

step 7

Next, we check for conditional convergence using the Alternating Series Test. The series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}}$ is an alternating series

step 8

For the Alternating Series Test, we need to check if $a_n = \frac{1}{1+\sqrt{n}}$ is decreasing and $\lim_{n \to \infty} a_n = 0$

step 9

The function $a_n = \frac{1}{1+\sqrt{n}}$ is decreasing because the denominator $1+\sqrt{n}$ is increasing. Also, $\lim_{n \to \infty} \frac{1}{1+\sqrt{n}} = 0$

step 10

Since both conditions of the Alternating Series Test are satisfied, the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}}$ converges conditionally

Answer

The series converges conditionally.

Key Concept

Alternating Series Test

Explanation

The series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}}$ converges conditionally because it satisfies the conditions of the Alternating Series Test: the terms decrease in magnitude and approach zero.

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