Problem 35 - Single Variable Calculus

Question

Find the value(s) of \(a\) for which the series \(\sum_{n=1}^{\infty}\left[\frac{a}{n+1}-\frac{1}{n+2}\right]\) converges. Justify your answer.

Step-by-Step Solution

Answer

The series $\sum_{n=1}^{\infty}\left[\frac{a}{n+1}-\frac{1}{n+2}\right]$ converges for all values of \(a\) except \(a=-2\).

Step 1: Rewrite the series

Let's first rewrite the given series more explicitly: \[\sum_{n=1}^{\infty}\left(\frac{a}{n+1}-\frac{1}{n+2}\right) = \frac{a}{2}-\frac{1}{3}+\frac{a}{3}-\frac{1}{4}+\frac{a}{4} - \frac{1}{5} + \cdots\]

Step 2: Apply the convergence test(s) and Telescoping sum

Notice that, the series can be written as a telescoping sum: \[\sum_{n=1}^{\infty}\left[\frac{a(n+2)-1}{(n+1)(n+2)}\right]\] Now, we can use partial fraction decomposition: \[\frac{a(n+2)-1}{(n+1)(n+2)} = \frac{A}{n+1}+\frac{B}{n+2}\] Multiplying both sides by \((n+1)(n+2)\), we have: \[a(n+2)-1 = A(n+2) + B(n+1)\] Comparing coefficients gives us the following system of linear equations: \[A+B = a\] \[2A + B = -1\] Now, we can solve for \(A\) and \(B\) in terms of \(a\): \[\begin{cases} A=2a+1\\ B=-a-2\end{cases}\] So, the series becomes: \[\sum_{n=1}^{\infty}\left(\frac{2a+1}{n+1}-\frac{a+2}{n+2}\right)\]

Step 3: Determine the value(s) of \(a\) for which the series converges

Since the series is now a telescoping sum, a vast majority of terms cancel out, and we are left with the sum of the first few terms: \[\lim_{N\to \infty}\left(\frac{2a+1}{1+1}-\frac{a+2}{N+2}\right)\] For the series to converge, this limit must be finite: \[\lim_{N\to \infty}\left(\frac{2a+1}{2}-\frac{a+2}{N+2}\right) < \infty\] This can only happen if the term \(\frac{a+2}{N+2}\) converges to 0 as \(N\) goes to infinity. In order for this to happen, we must have that \(a \neq -2\), as if \(a = -2\), the term would not converge to 0. Therefore, the series converges for all values of \(a\) except \(a = -2\).

Key Concepts

Telescoping Sum

A telescoping sum is a series where most terms cancel out when summed across a large number of terms. The example given involves a sequence of fractions: \( \sum_{n=1}^{\infty}\left(\frac{a}{n+1}-\frac{1}{n+2}\right) \). Generally, in a telescoping series, each term is canceled out by a subsequent term. This occurs because each fraction can be written such that they subtract from each other in sequence.When writing the series explicitly, you'll notice that each middle term cancels with a factor in the next fraction. For instance: - \( \frac{a}{2} - \frac{1}{3} + \frac{a}{3} - \frac{1}{4} + \ldots \) shows how most terms were eliminated.Ultimately, this means only the first and last terms can potentially contribute to the infinite sum.Understanding this cancelling nature is essential for identifying when a series simplifies significantly, hence aids in determining convergence.

Partial Fraction Decomposition

Partial fraction decomposition is the process of expressing a complex fraction as the sum of simpler fractions. This technique transforms expressions like \( \frac{a(n+2)-1}{(n+1)(n+2)} \) into parts that can be more easily analyzed.We set \( \frac{a(n+2)-1}{(n+1)(n+2)} = \frac{A}{n+1}+\frac{B}{n+2} \).
This allows us to calculate \( A \) and \( B \) by equating and comparing the numerators after multiplying both sides by the common denominator:

• First, set \( a(n+2) - 1 = A(n+2) + B(n+1) \)
• Solve for the system of linear equations obtained from comparing coefficients: \[ A + B = a \]
• \[ 2A + B = -1 \]

Through these equations, we determine \( A \) and \( B \) and break down the complex fraction to make it accessible for analysis and further steps, such as evaluating convergence.

Convergence Tests

Convergence tests help reveal if an infinite series will settle to a finite value or diverge. In this exercise, the transformed sum becomes:\( \sum_{n=1}^{\infty} \left(\frac{2a+1}{n+1} - \frac{a+2}{n+2}\right) \).We need this infinite series to eventually sum up to a fixed number. Various convergence tests, like the ratio or root test, can assist, but in telescoping sums, more emphasis is placed on the limiting behavior:- By observing how parts of the series behave separately, the focus shifts to whether individual terms cancel or converge to zero.In our context:

• The condition \( \lim_{N\to \infty}\frac{a+2}{N+2} = 0\) is crucial
• If \( a = -2 \), this term would cause issues as it wouldn't converge to zero

This clarity on convergence for all values except specific ones, like \( a = -2 \), emphasizes the series' behavior over infinity.

Limits and Infinity

When dealing with infinite series, understanding limits and infinity is essential as they define the behavior of a function or series as it approaches a large bound. The goal often is to ensure a series converges to a particular value:- The limit, \( \lim_{N\to \infty}\left(\frac{2a+1}{2} - \frac{a+2}{N+2}\right) \), captures the remaining sum as unsigned terms cancel.- For a series to converge, the impact of the 'end' of the series, represented by terms such as \( \frac{a+2}{N+2} \), must diminish as \( N \) becomes very large.Analyzing whether this limit maintains finiteness is critical:

• If the secondary term converges to zero — ideal in most cases — the remaining part maintains convergence
• Therefore, troubles arise when \( a = -2 \), since it prevents the denominator's term from approaching zero.

Understanding these aspects allows you to determine convergence and distinguish problematic values affecting the infinite series' behavior.