Math Problem Statement

so why is series 1/(n^2+1) divergent

Solution

To determine whether the series n=11n2+1\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} converges or diverges, we need to analyze the behavior of its terms and possibly compare it to a known convergent or divergent series.

One useful comparison is with the series n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2}, which is a well-known convergent series (the p-series with p=2p = 2).

Let's compare the terms of our series 1n2+1\frac{1}{n^2 + 1} with 1n2\frac{1}{n^2}:

1n2+1<1n2\frac{1}{n^2 + 1} < \frac{1}{n^2}

While 1n2+1\frac{1}{n^2 + 1} is less than 1n2\frac{1}{n^2}, this comparison alone does not help us determine the divergence or convergence because it does not satisfy the conditions needed for a comparison test to prove divergence. Instead, we should directly consider the behavior of the series itself.

We can use the Limit Comparison Test to compare 1n2+1\frac{1}{n^2 + 1} with 1n2\frac{1}{n^2}:

limn1n2+11n2=limnn2n2+1=limn11+1n2=1\lim_{{n \to \infty}} \frac{\frac{1}{n^2 + 1}}{\frac{1}{n^2}} = \lim_{{n \to \infty}} \frac{n^2}{n^2 + 1} = \lim_{{n \to \infty}} \frac{1}{1 + \frac{1}{n^2}} = 1

Since the limit is a finite, non-zero constant (in this case, 1), the Limit Comparison Test tells us that n=11n2+1\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} converges if and only if n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges. Because n=11n2\sum_{n=1}^{\infty} \frac{1}{n^2} is a convergent p-series with p=2>1p = 2 > 1, the series n=11n2+1\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} also converges.

Therefore, the series n=11n2+1\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} does not diverge; it converges.

Would you like further details or have any questions about this explanation?

Here are some more questions you might be interested in:

  1. Why does the series n=11n\sum_{n=1}^{\infty} \frac{1}{n} diverge?
  2. How do you determine if a series converges or diverges using the Ratio Test?
  3. What is the difference between absolute and conditional convergence?
  4. Can you explain the Alternating Series Test?
  5. How do you apply the Integral Test to determine the convergence of a series?

Tip: The Limit Comparison Test is a powerful tool for determining the convergence or divergence of a series by comparing it with another series whose behavior is already known.

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Math Problem Analysis

Mathematical Concepts

Series convergence
Limit comparison test
P-series

Formulas

Limit comparison test: \( \lim_{{n \to \infty}} \frac{a_n}{b_n} \)

Theorems

P-series convergence: \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \)

Suitable Grade Level

Advanced