Problem 48

Question

Which of the sequences $\left\\{a_{n}\right\\}$ converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=n \pi \cos (n \pi) $$

Step-by-Step Solution

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Answer

The sequence $a_n = n \pi \cos(n \pi)$ diverges.

1Step 1: Observing the Sequence Formula

The given sequence is defined as $ a_n = n \pi \cos(n \pi) $. Notice that $ \cos(n \pi) $ alternates between values, specifically $ \cos(n \pi) = (-1)^n $ for integers $ n $. This means that $ \cos(n \pi) $ is either $-1$ or $1$ depending on whether $n$ is odd or even.

2Step 2: Analyzing the Behavior of the Sequence

Substitute the property of $ \cos(n \pi) $ into the sequence formula: $ a_n = n \pi (-1)^n $. Therefore, the sequence alternates between $ n \pi $ and $-n \pi$. As $ n \to \infty $, both $ n \pi $ and $-n \pi$ go to $ \pm \infty $.

3Step 3: Determining Convergence or Divergence

Since $ a_n $ alternates between two unbounded expressions $ n\pi $ and $ -n\pi $, it does not settle towards any finite limit. This indicates the sequence is divergent.

Key Concepts

Alternating SequenceDivergent SequenceCosine FunctionInfinite Series

Alternating Sequence

An alternating sequence is a sequence in which the terms switch back and forth between specific values or patterns. In the context of this problem, the sequence is given by the formula $ a_n = n \pi \cos(n \pi) $. The term $ \cos(n \pi) $ alternates between $ -1 $ and $ 1 $, which means that every term in the sequence flips its sign based on whether $ n $ is odd or even. This creates a seesaw effect where the terms swing from one side to the other:

For even $ n $, $ \cos(n \pi) = 1 $
For odd $ n $, $ \cos(n \pi) = -1 $

When plugged back into the sequence $ a_n $, it results in terms that alternate between $ n \pi $ and $ -n \pi $, a behavior characteristic of alternating sequences.

Divergent Sequence

A divergent sequence is one that does not approach a specific value as the number of terms increases indefinitely. In our sequence $ a_n = n \pi \cos(n \pi) $, the divergence is evident because the terms alternate between $ n \pi $ and $ -n \pi $, increasing in magnitude as $ n $ becomes larger:

The terms do not hone in on a particular value; instead, they grow without bound.
For large $ n $, the terms move towards $ \pm \infty $ rather than stabilizing.

Because of this unbounded growth, we can conclude that the sequence is divergent. It does not have a limit.

Cosine Function

The cosine function, denoted as $ \cos(x) $, is a fundamental trigonometric function that relates to the coordinates of a point on a unit circle. In our sequence, $ \cos(n \pi) $ is a critical component due to its periodic property. On the unit circle:

$ \cos(x) $ reaches its maximum value of 1 when $ x = 0, \pm 2\pi, \pm 4\pi,... $
It reaches its minimum of -1 when $ x = \pm \pi, \pm 3\pi, \pm 5\pi,... $

For integer multiples of $ \pi $, $ \cos(n \pi) $ will either be -1 if $ n $ is odd or 1 if $ n $ is even. This oscillating pattern of the cosine function is what causes the sequence to alternate.

Infinite Series

An infinite series is a sum of infinitely many terms of a sequence. While our focus is on sequences, understanding infinite series can provide more depth when studying convergence. For our sequence $ a_n = n \pi \cos(n \pi) $, though not a series itself, it mirrors many properties you'd evaluate in series:

Terms of alternating sign do not guarantee convergence.
The absolute values of terms could grow without limit, as seen in diverging series.

In this case, each term in the sequence increases its magnitude indefinitely, representing a resemblance to what could be a divergent series. Understanding these concepts helps in evaluating whether a sequence or series has the potential to converge to a sum or not.

Problem 48

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