Question

Express the following series as a telescoping sum and evaluate its nth partial sum. $$ \sum_{n=1}^{\infty} \frac{(n+2)}{n(n+1) 2^{n+1}}\left(\text { Hint: Look at } 1 /\left(n 2^{n}\right) .\right) $$

Short answer

The series converges to 1.

Step-by-step solution

Decompose the Fraction

To begin, the given series can be tricky, but rewriting it in a more recognizable form helps. The fraction can be split using partial fraction decomposition. \[ \frac{n+2}{n(n+1)2^{n+1}} = \frac{A}{n} + \frac{B}{n+1} \] To find \(A\) and \(B\), equate: \( n+2 = A(n+1) + Bn \). Solving for \(A\) and \(B\), we find \(A=1\), \(B=1\). Hence, \[ \frac{n+2}{n(n+1)2^{n+1}} = \frac{1}{n2^{n+1}} - \frac{1}{(n+1)2^{n+1}} \].

Recognize the Telescoping Nature

Now, notice the terms \( \frac{1}{n2^{n+1}} - \frac{1}{(n+1)2^{n+1}} \) create a telescoping effect. A telescoping series is one where most terms cancel each other. Begin listing the first few terms based on the partial fractions.

Evaluate the nth Partial Sum

The nth partial sum of our series is: \[ S_n = \sum_{k=1}^{n} \left(\frac{1}{k2^{k+1}} - \frac{1}{(k+1)2^{k+1}}\right) \]. Due to its telescoping nature, most terms cancel. What remains is: \[ S_n = 1 - \frac{1}{(n+1) 2^{n+1}} \].

Express the Sum as n approaches Infinity

As \( n \to \infty \), the term \( \frac{1}{(n+1)2^{n+1}} \to 0 \). Therefore, the sum of the series as \( n \to \infty \) is \[ S = 1 \].

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions, which are often more manageable to work with in series or integrals. In the exercise, we started with the fraction \( \frac{n+2}{n(n+1)2^{n+1}} \). To decompose this, we express it as a sum of simpler fractions like \( \frac{A}{n} + \frac{B}{n+1} \).

This process involves the following steps:

• Express the original fraction in terms of unknown constants, here \(A\) and \(B\).
• Equate the numerators after multiplying through by the common denominator \(n(n+1)\).
• Solve the resulting system of equations for the unknowns \(A\) and \(B\).

Once this decomposition is done, it facilitates recognizing patterns, such as telescoping, in a series. Here, it allowed us to rewrite the sequence in a way that terms could easily cancel out, which is critical in evaluating the series efficiently.

Partial Sum

Partial sums are foundational when dealing with infinite series, including telescoping series. The partial sum \( S_n \) represents the sum of the first \( n \) terms in a series. In the exercise, this was expressed as:\[ S_n = \sum_{k=1}^{n} \left(\frac{1}{k2^{k+1}} - \frac{1}{(k+1)2^{k+1}}\right) \]The partial sum \( S_n \) is crucial in series analysis as it allows us to see how the series builds term by term.
By calculating partial sums, we can observe patterns, identify telescoping nature, and determine behavior as \( n \) approaches infinity. In telescoping sums, a series' partial terms cancel out with subsequent or preceding terms, making the process of evaluation straightforward, often boiling down to the first and last terms in the sequence.

Infinite Series

An infinite series is the sum of an infinite sequence, often expressed in the form \( \sum_{n=1}^{\infty} a_n \). In our particular exercise, the infinite series initially seems daunting due to its complexity:

\[ \sum_{n=1}^{\infty} \frac{(n+2)}{n(n+1) 2^{n+1}} \]

After partial fraction decomposition, the series simplifies into a telescoping series. Infinite series can converge to a finite value or diverge depending on the properties of the sequence involved. Transforming complex series into a simpler form often reveals whether a sum can be computed or approximated as \( n \to \infty \).

• For non-specialists, being able to rewrite an infinite series is crucial as it offers insight on applying tests for convergence or establishing sums.
• Whether dealing with power series, geometric series, or telescoping ones like in this exercise, understanding its behavior as terms keep adding is key.

Convergence of Series

Convergence is a core concept in series analysis, determining the eventual sum of a series as its terms proceed to infinity. In this exercise, the nature of convergence was explored after breaking the series into a telescoping sum. We determined the infinite series sum by considering what happens as \( n \) approaches infinity.

The nth partial sum \( S_n = 1 - \frac{1}{(n+1) 2^{n+1}} \) shows that as \( n \to \infty \), the term \( \frac{1}{(n+1) 2^{n+1}} \) diminishes to zero. This leads to the conclusion that the series converges to 1. This convergence is because the terms that remain as \( n \) increases have negligible contributions to the sum.

• Understanding convergence involves knowing several series tests, such as the ratio test, integral test, and others, to evaluate whether partial sums approach a finite limit.
• Convergence plays a vital role, as it distinguishes whether a sequence of numbers essentially stabilizes to a sum or continues indefinitely without settling.

Recognizing series convergence is critical when gauging the behavior of functions and sequences in broader mathematical applications.