Question

Find the function represented by the following series and find the interval of convergence of the series. $$\sum_{k=0}^{\infty}\left(x^{2}+1\right)^{2 k}$$

Short answer

Answer: The function represented by the given geometric series is $$S(x) = \frac{1}{1-\left(x^{2}+1\right)^{2}}$$, and the interval of convergence is empty, which means the series does not converge for any real value of \(x\).

Step-by-step solution

Recognize the given series as a geometric series

Notice that the given series is a geometric series with the first term as \(1\) (when \(k=0\)) and the common ratio as \(\left(x^{2}+1\right)^{2}\).

Write the formula for the sum of a geometric series

The formula for the sum of a geometric series with the first term \(a\) and the common ratio \(r\) is as follows: $$S(x) = \frac{a}{1-r}$$ In our case, \(a=1\) and \(r=\left(x^{2}+1\right)^{2}\). Thus, the formula for the sum of our series is: $$S(x) = \frac{1}{1-\left(x^{2}+1\right)^{2}}$$

Determine the interval of convergence

A geometric series converges if the common ratio has an absolute value less than \(1\). In our case, the common ratio is: $$\left|\left(x^{2}+1\right)^{2}\right|$$ So, the series converges if: $$0 < \left|\left(x^{2}+1\right)^{2}\right| < 1$$ Now, we know that \((x^2 + 1)^2 \geq 0\) for all values of \(x\). So, the left inequality is satisfied for all real numbers. To find the interval of convergence, we need to satisfy the right inequality: $$\left|\left(x^{2}+1\right)^{2}\right| < 1$$ Notice that if the absolute value of a square is less than \(1\), then the original expression itself must be less than \(1\): $$\left(x^{2}+1\right)^{2} < 1$$ Taking the square root of both sides, we get: $$x^2+1 < 1$$ $$x^2 < 0$$ However, there's no real number whose square is less than \(0\). Hence, the given series does not converge for any real value of \(x\). In summary, the function represented by the given series is: $$S(x) = \frac{1}{1-\left(x^{2}+1\right)^{2}}$$ And the interval of convergence is empty.

Key concepts

Convergence Criteria for Geometric Series

To determine if a geometric series converges, we employ specific criteria. A geometric series is a series of the form \(a + ar + ar^2 + ar^3 + \ldots \), where \(a\) is the first term, and \(r\) is the common ratio. This series converges absolutely if the absolute value of the common ratio, \(|r|\), is less than 1. When \(|r| < 1\), the infinite sum of the series approaches a finite value, ensuring convergence.

However, when \(|r| \geq 1\), the series does not converge, meaning the sum could become infinitely large or simply doesn't settle to a fixed number. In the exercise, the common ratio \(\left(x^{2}+1\right)^{2}\) had a magnitude greater than or equal to 1 for all real \(x\), indicating that our series did not converge for any real values of \(x\). Understanding this convergence criterion is crucial in studying the behavior of infinite series to determine when sums make sense or are meaningful.

Series Summation Formula in a Geometric Series

The formula to find the sum of an infinite geometric series is simple and elegant. Given a series with the first term \(a\) and a common ratio \(r\), the sum \(S\) is expressed as \(S = \frac{a}{1-r}\), provided that \(|r| < 1\). This formula gives us a shortcut to calculate the total sum directly without adding up each term individually over an infinite series.

For example, in the given exercise, the first term \(a = 1\) and the common ratio \(r = \left(x^{2}+1\right)^{2}\). Plugging these values into the formula results in \(S(x) = \frac{1}{1 - \left(x^{2}+1\right)^{2}}\). This represents the function of our geometric series. The catch here, however, is that we need \(|r| < 1\) for this formula to be valid. If this condition fails, the formula no longer represents a converging sum.

Interval of Convergence

The interval of convergence in a series determines the set of values for which the series sum converges to a single value. For a geometric series, once we have the common ratio \(r\), we verify through \(|r| < 1\) to determine where the series converges.

In the exercise, the expression \((x^2 + 1)^2 < 1\) was key to finding this interval. Solving this inequality brought us to a contradiction, \(x^2 < 0\). In simple terms, there are no real numbers that satisfy this because squaring a real number never results in a negative. Hence, the interval of convergence was deemed empty, meaning the series converges for no real values of \(x\). Understanding and determining the interval of convergence helps in figuring out the behavior of the function represented by the series over different inputs.