Chapter 11: Problem 13
Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why. $$ \sum_{n=1}^{\infty} \frac{1}{n^{2}+1} $$
Short Answer
Step by step solution
Understand and Write the Series
Calculate the First 10 Terms
Calculate Partial Sums
Graph the Terms and Partial Sums
Analyze the Convergence
Determine Convergence or Divergence
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Sums
- First partial sum \( S_1 = a_1 \)
- Second partial sum \( S_2 = a_1 + a_2 \)
- ... and so on ...
In the original exercise, you calculated the first ten partial sums to inspect the sequence's behavior. As you add each term (\( \rac{1}{2}, \rac{1}{5}, \rac{1}{10} \), etc.), the cumulative sum gives an insight into the nature of convergence. These partial sums, when plotted, visually depict whether they approach a certain limit.
Sequences and Series
There are two main types of series: convergent, where the sum approaches a finite limit, and divergent, where the sum grows without bound. Recognizing whether a series converges or diverges is key to analyzing its behavior. Tools such as partial sums offer a practical approach to examine this, while advanced tests can confirm the nature of a series definitively.
In your series, each term gets smaller as \( n \) increases, hinting towards convergence. However, to affirm this, you must perform further analysis or apply specific convergence tests.
Comparison Test
- If a series with positive terms is less than a well-known convergent series, it too converges.
- Conversely, if it is greater than a divergent series, it diverges.
In the given series \( \sum_{n=1}^{\infty} \frac{1}{n^2+1} \), the terms \( \frac{1}{n^2+1} \) behave similarly to the terms of the well-known \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is convergent because the exponent \( p \) is greater than 1. Since \( \frac{1}{n^2+1} \) is always less than \( \frac{1}{n^2} \), the comparison test confirms convergence.
This logic shows that, like the \( p \)-series, the terms of your given series will sum to a finite value, supporting the conclusion that the series converges.