Final answer:
The sequence aₙ = (⅒/n + 1)(-1ⁿ) is bounded as it oscillates between -1 and 1, and the magnitude of each term does not exceed 1, thus a practical bound can be set at M = 2.
Step-by-step explanation:
To understand if the sequence aₙ = (⅒/n + 1)(-1ⁿ) is bounded, we examine the behavior of the sequence terms as n approaches infinity.
The sequence alternates signs because of the (-1ⁿ) term. However, the magnitude of each term is governed by the (⅒/n + 1) part. As n increases, the term 1/n approaches zero, and thus each term of the sequence approaches 1 as n approaches infinity, albeit with alternating signs.
A sequence is bounded if there is a real number M such that the absolute value of every term in the sequence is less than or equal to M
. Since the sequence aₙ consists of factors that approach 1 in magnitude and simply alternate in sign, it is clear that the sequence is indeed bounded. A practical bound for this sequence is M = 2, since none of the sequence terms will exceed this value in magnitude.
We can conclude that aₙ = (⅒/n + 1)(-1ⁿ) is a bounded sequence because it oscillates between -1 and 1, never exceeding these values in magnitude.