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In mathematics, an oscillating sequence is one where the terms do not settle to a single value. Instead, they continually go back and forth between different values. Such sequences never converge because they do not approach a particular number. A good example of this is the sequence given by \((-1)^n\), which oscillates between \-1\ and \1\. The oscillating behavior results from the power of \-1\, which flips the sign for consecutive terms:

• When \ n \ is even, \ (-1)^n = 1 \.
• When \ n \ is odd, \ (-1)^n = -1 \.

The alternating sign means the sequence itself doesn't settle into a routine pattern or converge to a single value. However, by pairing this oscillating sequence as a numerator over a growing denominator, \( n \), as seen in the exercise, the influence of the oscillation diminishes as the denominator grows larger.

When discussing the limit of a sequence as \( n \) approaches infinity, it refers to the behavior of the sequence as the term number gets very large. In our exercise, the limit we are considering is \( \lim_{n \to \infty} \frac{(-1)^n}{n} \). To understand this, knowing how both the numerator and the denominator behave is crucial.The denominator \( n \) grows infinitely large. An infinite limit suggests that the terms of the sequence may grow beyond bound, but in this specific case, the oscillating numerator means the sequence’s growth diverges only in proportion to changes in sign, not magnitude. Instead, the value of each fraction diminishes:

• Even when the numerator fluctuates between \-1\ and \1\, an infinitely large denominator causes each individual term to shrink closer to zero.
• The larger the denominator, the smaller the overall value of each term becomes.

This property helps take even sequences with non-convergent numerators, like those that oscillate, towards convergence to a finite limit.

The concept of convergence in sequences is fundamental in understanding limits. A sequence is said to converge if its terms approach a single, finite value as \( n \) becomes very large.In our exercise, the sequence \ \frac{(-1)^n}{n} \ converges to zero. Here's why:

• Even though \ (-1)^n \ oscillates and doesn't converge on its own, the denominator \ n \ grows without bound.
• The result of \ \frac{(-1)^n}{n} \ becomes negligently small because each term is divided by a larger and larger number, overpowering the oscillation in the numerator. Both \ \frac{1}{n} \ and \ -\frac{1}{n} \ tend towards zero as \( n \) increases.

Thus, the sequence ultimately converges to zero. This illustrates the importance of analyzing both the numerator and the denominator to understand convergence in sequences.