Question

Evaluate the following telescoping series or state whether the series diverges. $$ \sum_{n=1}^{\infty} \frac{1}{n^{13}}-\frac{1}{(n+1)^{12}} $$

Short answer

The series converges.

Step-by-step solution

Identify the terms of the sequence

The series given is \(\sum_{n=1}^{\infty} \left( \frac{1}{n^{13}} - \frac{1}{(n+1)^{12}} \right)\). It consists of two main expressions: \(\frac{1}{n^{13}}\) and \(-\frac{1}{(n+1)^{12}}\).

Analyze the series type

This is a telescoping series, which typically simplifies due to cancellation of consecutive terms. Let's examine how these terms may cancel by writing out the first few terms explicitly.

Write out initial terms

Substitute initial values of \(n\) into the series to see the pattern:- For \(n=1\), \(\frac{1}{1^{13}} - \frac{1}{2^{12}}\)- For \(n=2\), \(\frac{1}{2^{13}} - \frac{1}{3^{12}}\)- For \(n=3\), \(\frac{1}{3^{13}} - \frac{1}{4^{12}}\)Observe that terms do not easily cancel in a telescoping manner due to different exponents.

Check convergence

Since simple cancellation doesn't occur, evaluate if individual sequence terms lead to convergence. The series can be viewed as two separate series:\( \sum_{n=1}^{\infty} \frac{1}{n^{13}} \) – convergent because p-series with \(p=13>1\).\( \sum_{n=1}^{\infty} \frac{1}{(n+1)^{12}} \) – also convergent because p-series with \(p=12>1\).

Key concepts

Convergence of Series

When we talk about the convergence of a series, we are interested in whether the sum of an infinite sequence of numbers approaches a specific value. If it does, the series converges; if not, it diverges. Determining the convergence of a series is crucial because it tells us if the series behaves in a predictable way.

To check convergence, various tests can be applied depending on the series type. In our case, we have a telescoping series where cancellation usually occurs among terms. However, this example did not simplify as traditionally expected because the terms had different exponents. Nonetheless, assessing convergence is still possible by treating it as two separate series.

• The first series: \[ \sum_{n=1}^{\infty} \frac{1}{n^{13}} \]
• The second series: \[ \sum_{n=1}^{\infty} \frac{1}{(n+1)^{12}} \]

Each of these series converges individually due to their nature as p-series with exponents greater than 1. Therefore, the combination of their behaviors helps determine how the original series acts. Understanding these individual behaviors is key in assessing the overall convergence.

P-Series

A p-series takes the form:\[sum_{n=1}^{\infty} \frac{1}{n^p}\]where \(p\) is a constant that affects convergence. P-Series are a fundamental concept in evaluating the convergence of series.

P-series convergence is straightforward:

• If \(p > 1\), then the series converges.
• If \(p \leq 1\), then the series diverges.

The simplicity of this test makes it greatly useful in many problems.

In our exercise, each portion of the series can be viewed as a p-series:

• For \(\frac{1}{n^{13}}\), \(p = 13\), so it converges because \(p > 1\).
• For \(\frac{1}{(n+1)^{12}}\), \(p = 12\), so it also converges since \(p > 1\).

This understanding allows us to assert the convergence of more complex series like the given one by analyzing its components using the p-series test.

Series Evaluation

Evaluating series generally involves adding an infinite set of numbers. It can sometimes be tricky because you need methods to handle infinite terms. Series like those discussed involve concepts such as telescoping and p-series to work out a solution.

Telescoping series are series in which partial cancellations make evaluation straightforward, revealing a simple sum. In this exercise, however, direct cancellation was not evident.

When simple telescoping isn't an option, breaking down the series into components that can be analyzed independently often helps. In this case, although the given series did not telescope cleanly, we evaluated it by assessing each component's convergence as a p-series.

• Identified the series as a sum of two separate p-series.
• Determined the convergence by applying the p-series rule.

Series evaluation requires these techniques to effectively deal with their infinite nature and to determine their convergence or divergence.