Infinite Series Summation And Convergence Analysis Of $\sum_{n=1}^{\infty} \frac{12}{n(n+3)}$

We delve into the fascinating realm of infinite series, specifically focusing on the series represented by the summation: $\sum_{n=1}^{\infty} \frac{12}{n(n+3)}$. This exploration will involve both analytical techniques to determine the sum of the series and computational methods to approximate the sum using partial sums. Our journey will encompass the following key aspects:

• (a) Analytical Determination of the Sum: Employing partial fraction decomposition and telescoping series techniques to derive the exact sum of the infinite series, rounded to four decimal places.
• (b) Numerical Approximation using Partial Sums: Utilizing a graphing utility or computational software to calculate indicated partial sums ($S_n$) for various values of n and observing the convergence behavior of the series.
• Discussion and Interpretation: Analyzing the results obtained from both analytical and numerical approaches, discussing the convergence properties of the series, and highlighting the relationship between partial sums and the infinite sum.

(a) Finding the Sum of the Series: A Journey Through Partial Fractions and Telescoping Sums

Our quest to find the sum of the infinite series $\sum_{n=1}^{\infty} \frac{12}{n(n+3)}$ begins with the crucial technique of partial fraction decomposition. This method allows us to break down the complex rational expression $\frac{12}{n(n+3)}$ into simpler fractions that are easier to work with. The essence of partial fraction decomposition lies in expressing the given fraction as a sum of fractions with linear denominators. In this case, we seek constants A and B such that:

$\frac{12}{n(n+3)} = \frac{A}{n} + \frac{B}{n+3}$

To determine the values of A and B, we multiply both sides of the equation by the common denominator n(n+3), resulting in:

$12 = A(n+3) + Bn$

Expanding the equation, we get:

$12 = An + 3A + Bn$

Now, we can group the terms with n and the constant terms:

$12 = (A+B)n + 3A$

For this equation to hold true for all values of n, the coefficients of n on both sides must be equal, and the constant terms must be equal. This leads to the following system of equations:

$A + B = 0$

$3A = 12$

From the second equation, we can directly solve for A:

$A = \frac{12}{3} = 4$

Substituting the value of A into the first equation, we can solve for B:

$4 + B = 0$

$B = -4$

Thus, we have successfully decomposed the fraction into partial fractions:

$\frac{12}{n(n+3)} = \frac{4}{n} - \frac{4}{n+3}$

Now, we can rewrite the original series using this decomposition:

$\sum_{n=1}^{\infty} \frac{12}{n(n+3)} = \sum_{n=1}^{\infty} \left( \frac{4}{n} - \frac{4}{n+3} \right)$

The beauty of this form lies in its telescoping nature. A telescoping series is a series where most of the terms cancel out, leaving only a few terms at the beginning and end. To see this telescoping behavior, let's write out the first few terms of the series:

$\left( \frac{4}{1} - \frac{4}{4} \right) + \left( \frac{4}{2} - \frac{4}{5} \right) + \left( \frac{4}{3} - \frac{4}{6} \right) + \left( \frac{4}{4} - \frac{4}{7} \right) + \left( \frac{4}{5} - \frac{4}{8} \right) + \left( \frac{4}{6} - \frac{4}{9} \right) + ...$

Notice how the $-\frac{4}{4}$ term cancels with the $\frac{4}{4}$ term, the $-\frac{4}{5}$ term cancels with the $\frac{4}{5}$ term, and so on. This cancellation pattern continues indefinitely. To find the sum of the infinite series, we consider the n-th partial sum, denoted by $S_n$:

$S_n = \sum_{k=1}^{n} \left( \frac{4}{k} - \frac{4}{k+3} \right)$

Writing out the terms of the partial sum, we observe the cancellations:

$S_n = \left( \frac{4}{1} - \frac{4}{4} \right) + \left( \frac{4}{2} - \frac{4}{5} \right) + \left( \frac{4}{3} - \frac{4}{6} \right) + ... + \left( \frac{4}{n-2} - \frac{4}{n+1} \right) + \left( \frac{4}{n-1} - \frac{4}{n+2} \right) + \left( \frac{4}{n} - \frac{4}{n+3} \right)$

After the cancellations, we are left with:

$S_n = \frac{4}{1} + \frac{4}{2} + \frac{4}{3} - \frac{4}{n+1} - \frac{4}{n+2} - \frac{4}{n+3}$

Now, to find the sum of the infinite series, we take the limit of the partial sum as n approaches infinity:

$\sum_{n=1}^{\infty} \frac{12}{n(n+3)} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{4}{1} + \frac{4}{2} + \frac{4}{3} - \frac{4}{n+1} - \frac{4}{n+2} - \frac{4}{n+3} \right)$

As n approaches infinity, the terms $\frac{4}{n+1}$, $\frac{4}{n+2}$, and $\frac{4}{n+3}$ approach zero. Therefore, the sum of the infinite series is:

$\sum_{n=1}^{\infty} \frac{12}{n(n+3)} = 4 + 2 + \frac{4}{3} = 6 + \frac{4}{3} = \frac{18}{3} + \frac{4}{3} = \frac{22}{3}$

Converting this fraction to a decimal and rounding to four decimal places, we get:

$\frac{22}{3} \approx 7.3333$

Thus, the sum of the series is approximately 7.3333. This analytical approach, employing partial fraction decomposition and the concept of telescoping series, has allowed us to precisely determine the sum of the given infinite series.

(b) Numerical Approximation using Partial Sums: Exploring Convergence with a Graphing Utility

In this section, we transition from the analytical determination of the series sum to a numerical approximation using partial sums. This approach involves calculating the sum of the first n terms of the series, denoted as $S_n$, for various values of n. By observing the behavior of $S_n$ as n increases, we can gain insights into the convergence of the series and approximate its sum.

To facilitate this process, we employ a graphing utility or computational software. These tools provide the capability to efficiently compute partial sums for a wide range of n values. We will calculate $S_n$ for several values of n, such as n = 10, 50, 100, 500, and 1000, and observe the trend in the values of $S_n$.

Let's define the n-th partial sum as:

$S_n = \sum_{k=1}^{n} \frac{12}{k(k+3)}$

Using the partial fraction decomposition we derived earlier, we can rewrite the partial sum as:

$S_n = \sum_{k=1}^{n} \left( \frac{4}{k} - \frac{4}{k+3} \right) = 4 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+3} \right)$

This form is computationally advantageous as it simplifies the calculation of the partial sums. Now, let's compute the partial sums for the specified values of n:

• For n = 10:

  $S_{10} = 4 \sum_{k=1}^{10} \left( \frac{1}{k} - \frac{1}{k+3} \right) \approx 6.9254$

• For n = 50:

  $S_{50} = 4 \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+3} \right) \approx 7.2681$

• For n = 100:

  $S_{100} = 4 \sum_{k=1}^{100} \left( \frac{1}{k} - \frac{1}{k+3} \right) \approx 7.3037$

• For n = 500:

  $S_{500} = 4 \sum_{k=1}^{500} \left( \frac{1}{k} - \frac{1}{k+3} \right) \approx 7.3267$

• For n = 1000:

  $S_{1000} = 4 \sum_{k=1}^{1000} \left( \frac{1}{k} - \frac{1}{k+3} \right) \approx 7.3299$

Observing these values, we notice a clear trend: as n increases, the partial sums $S_n$ approach a certain value. This behavior is indicative of the series converging to a finite sum. The values of $S_n$ for larger n (e.g., n = 500 and n = 1000) are close to the analytical result we obtained earlier (7.3333), further supporting the convergence of the series and validating our analytical calculation.

To visualize this convergence, we can plot the partial sums $S_n$ as a function of n. The plot would show the partial sums approaching a horizontal asymptote, which represents the sum of the infinite series. This graphical representation provides a compelling illustration of the convergence behavior.

The numerical approximation using partial sums complements the analytical approach by providing a computational perspective on the series convergence. It allows us to observe the gradual accumulation of terms and the approach towards the limiting sum. This method is particularly useful for series where an analytical solution may be difficult or impossible to obtain.

Discussion and Interpretation: Bridging Analytical and Numerical Perspectives

Our exploration of the series $\sum_{n=1}^{\infty} \frac{12}{n(n+3)}$ has yielded valuable insights through both analytical and numerical approaches. The analytical method, employing partial fraction decomposition and telescoping series, provided us with the exact sum of the series, which we found to be $\frac{22}{3} \approx 7.3333$. The numerical method, using partial sums calculated with a graphing utility, allowed us to approximate the sum and observe the convergence behavior of the series.

The consistency between the analytical result and the numerical approximations reinforces our understanding of the series. The partial sums, as n increased, progressively approached the analytical sum, demonstrating the convergence of the series. This convergence is a fundamental property of the series, indicating that the sum of infinitely many terms approaches a finite value.

The telescoping nature of the series, revealed through partial fraction decomposition, played a crucial role in determining the sum analytically. The cancellations of terms in the partial sums simplified the expression and allowed us to easily evaluate the limit as n approached infinity. This technique is a powerful tool for dealing with series of this form.

The numerical approximation method provides a practical way to estimate the sum of a series, especially when an analytical solution is not readily available. By calculating partial sums for sufficiently large n, we can obtain a close approximation to the infinite sum. The accuracy of the approximation improves as n increases, highlighting the importance of considering a large number of terms for a reliable estimate.

Furthermore, the graphing utility enabled us to visualize the convergence of the series. Plotting the partial sums as a function of n revealed a clear trend towards a horizontal asymptote, representing the sum of the series. This visual representation enhances our understanding of the convergence process and provides a complementary perspective to the numerical data.

In conclusion, our comprehensive analysis of the series $\sum_{n=1}^{\infty} \frac{12}{n(n+3)}$ has showcased the power of both analytical and numerical techniques. The combination of these approaches has provided a deep understanding of the series' behavior, its convergence properties, and its sum. This exploration serves as a valuable example of how different mathematical tools can be used in concert to solve problems and gain insights into the fascinating world of infinite series.