Question

Evaluate \(\sum_{n=2}^{\infty} \frac{2}{n^{3}-n}\)

Short answer

The series converges to 1.

Step-by-step solution

Simplify the General Term

Observe that the denominator of the general term is a difference of squares: \(n^3 - n = n(n^2 - 1) = n(n-1)(n+1)\). This suggests the possibility of using partial fraction decomposition to simplify each term of the series.

Perform Partial Fraction Decomposition

Express \(\frac{2}{n(n-1)(n+1)}\) as:\[\frac{2}{n(n-1)(n+1)} = \frac{A}{n} + \frac{B}{n-1} + \frac{C}{n+1}\]Multiply through by the denominator \(n(n-1)(n+1)\) to obtain:\[2 = A(n-1)(n+1) + Bn(n+1) + Cn(n-1)\] Determine constants \(A, B,\) and \(C\) by choosing convenient values for \(n\) or by equating coefficients.

Solve for Constants

By substituting convenient values of \(n\):- Let \(n = 0\): \(2 = -A\), so \(A = -2\).- Let \(n = 1\): \(2 = B(1 \cdot 2)\), so \(B = 1\).- Let \(n = -1\): \(2 = C((-1)(-2))\), so \(C = 1\).Thus, \(A = -2, B = 1, C = 1\). Substitute these back into the partial fraction form:

Substitution into the Series

The original expression becomes:\[\sum_{n=2}^{\infty} \left (-\frac{2}{n} + \frac{1}{n-1} + \frac{1}{n+1} \right)\]

Apply Series Rewriting and Cancellation

Rewrite the series as:\[\sum_{n=2}^{\infty} \left(\frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1}\right)\]Notice that this is a telescoping series. Rewrite terms and observe cancellation when terms are unpacked.

Evaluate the Telescoping Series

When expanding the series, terms will cancel out:- \(\frac{1}{1} + (\frac{1}{2} - \frac{2}{2} + \frac{1}{3}) + (\frac{1}{3} - \frac{2}{3} + \frac{1}{4}) + \ldots\)After cancellation, you are left with the first term (\(\frac{1}{1} = 1\)) and remaining tail terms. As \(n \to \infty\), the series converges to \(1\).

Conclusion

After observing that subsequent terms continue to cancel beyond what is explicitly written, leaving a remainder of \(\frac{1}{1}\), you conclude:The series converges to 1.

Key concepts

Partial Fraction Decomposition

In mathematics, partial fraction decomposition is a technique used to simplify complex rational expressions. It involves breaking down a fraction into simpler parts called partial fractions. This method is particularly useful when dealing with polynomials in the denominator that can be factored into linear terms.
To illustrate, consider the rational expression \( \frac{2}{n(n-1)(n+1)} \). Here, partial fraction decomposition works by expressing this fraction as a sum of simpler fractions:

• \(\frac{A}{n}\)
• \(\frac{B}{n-1}\)
• \(\frac{C}{n+1}\)

The next step involves solving for the constants \(A\), \(B\), and \(C\) by applying techniques such as choosing strategic values for \(n\) or equating coefficients. This step is crucial as it allows for the original expression to be rewritten in a format that is easier to manipulate or analyze. In this specific case, solving the equations will give \(A = -2\), \(B = 1\), and \(C = 1\).

Telescoping Series

A telescoping series is a series where most terms cancel each other out when the series is expanded. This property simplifies the task of finding the series sum, as many intermediate terms vanish, leaving only a few terms. The series then effectively 'telescopes' down to a more manageable form.
In this exercise, we encounter a telescoping series after applying partial fraction decomposition. The series:

• \(\sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1} \right)\)

is telescoping because its terms begin to cancel each other in the expansion:
For instance, the \( \frac{1}{n-1} \) term cancels with part of a later term, leaving fewer terms to consider. This pattern continues as more terms are considered from the series expansion.

Convergence of Series

Convergence is an essential concept in infinite series. It determines whether an infinite series has a finite sum or not. For a series to converge, the infinite addition of its terms must approach a specific, finite value.
In this problem, the partial sum:

• \(1 + (\frac{1}{2} - \frac{2}{2} + \frac{1}{3}) + (\frac{1}{3} - \frac{2}{3} + \frac{1}{4}) + \ldots \)

illustrates the telescoping nature of the series, where several terms cancel out, particularly those beyond the scope of direct observation. As the number of terms approaches infinity, the calculation reveals that all but the \(\frac{1}{1}\) term cancel out completely. Thus, the series converges to the value 1, signifying that the infinite sum results in a finite, specific number.

Series Manipulation

Series manipulation involves the act of rewriting and rearranging terms in a series to simplify evaluation. This might include factoring denominators, reordering terms, or applying different mathematical identities.
For this series:

• The series was initially transformed using partial fraction decomposition.
• Then it was arranged as a telescoping series to aid in the cancellation of terms.
• Finally, terms were systematically grouped and evaluated.

By carefully rearranging and applying specific operations, the complex problem becomes easier to handle and the solution more apparent. These manipulations are crucial in series evaluation as they allow the identification of hidden patterns, such as telescoping, that lead to convergence insights and simplify the calculation of sums.