Math Problem Statement

it is real analysis questions

1. Show that \(\sum_{n=1}^{\infty} \frac{1}{n^2} < \infty\) (i.e., the series is convergent). 2. Show that if \(\sum_{n=1}^{\infty} |a_n| < \infty\), then \(\sum_{n=0}^{\infty} a_n\) is convergent.

Solution

Let's tackle each question in turn.

Question 1

We need to show that the series n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges.

Solution:

This series is a p-series, which is given by n=11np\sum_{n=1}^{\infty} \frac{1}{n^p}, where pp is a positive constant. The convergence of a p-series depends on the value of pp:

  • If p>1p > 1, the series converges.
  • If 0<p10 < p \leq 1, the series diverges.

In our case: p=2>1p = 2 > 1 Therefore, n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges.

To formally confirm this, we can compare it to the integral test for convergence. The integral test states that for a positive, decreasing, continuous function f(x)=1x2f(x) = \frac{1}{x^2}, the convergence of the integral 11x2dx\int_1^{\infty} \frac{1}{x^2} \, dx implies the convergence of the series n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}.

Calculating the integral: 11x2dx=[1x]1=1\int_1^{\infty} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^{\infty} = 1 Since the integral converges, by the integral test, the series n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges as well.

Question 2

We need to show that if n=1an<\sum_{n=1}^{\infty} |a_n| < \infty, then n=0an\sum_{n=0}^{\infty} a_n is convergent.

Solution:

This question is about the absolute convergence of a series.

  1. Absolute Convergence Implies Convergence: If n=1an\sum_{n=1}^{\infty} |a_n| converges, we say that the series n=1an\sum_{n=1}^{\infty} a_n is absolutely convergent.
  2. A well-known result in real analysis states that if a series is absolutely convergent, then it is also convergent.

Therefore, if n=1an\sum_{n=1}^{\infty} |a_n| converges, it implies that n=1an\sum_{n=1}^{\infty} a_n also converges.

Summary

  1. For Question 1, we showed that n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges because it is a p-series with p=2>1p = 2 > 1.
  2. For Question 2, we applied the fact that absolute convergence implies convergence, so if n=1an\sum_{n=1}^{\infty} |a_n| converges, then n=0an\sum_{n=0}^{\infty} a_n also converges.

Would you like further details on any of these steps, or do you have additional questions?

Related Questions

  1. What is the difference between conditional and absolute convergence?
  2. How does the comparison test work in determining the convergence of a series?
  3. Can the integral test be used to prove divergence as well as convergence?
  4. What are some examples of p-series that diverge?
  5. How does the ratio test apply to absolute convergence?

Tip

When analyzing series, always check if it's a well-known type (like geometric or p-series) first, as this can often immediately reveal its convergence properties.

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Math Problem Analysis

Mathematical Concepts

Series convergence
p-series
Absolute convergence
Integral test

Formulas

\(\sum_{n=1}^{\infty} \frac{1}{n^p}\) for p-series convergence
\(\int_1^{\infty} \frac{1}{x^p} \, dx\) for the integral test

Theorems

p-series test
Integral test
Absolute convergence implies convergence theorem

Suitable Grade Level

Undergraduate Mathematics