Problem 13 - Advanced Calculus

Question

Prove that \(\lim _{n \rightarrow x} c^{1 / n}=1\) for any \(c>1\) by setting \(a_{n}=c^{1 / n}-1\), and then deriving the estimate \(0 \leq a_{n} \leq(c-1) / n\)

Step-by-Step Solution

Answer

The limit \(\lim_{n \rightarrow \infty}c^{1/n} = 1\) for all \(c>1\) is proved by setting \(a_n = c^{1/n} - 1\), showing that \(a_n\) is decreasing and bounded, thereby allowing us to use the Monotone Convergence Theorem to establish its limit as \(n\) approaches infinity.

Step 1: Identify the given sequence

We are given a sequence \(a_{n}=c^{1 / n}-1\), where \(c\) is a constant greater than 1

Step 2: Determine the monotonocity of the sequence

We need to prove that the sequence \(a_{n}\) is decreasing. This can be done by showing that \(a_{n+1}\leq a_{n}\) for all \(n\). By definition of \(a_{n}\), we have \(a_{n+1}=c^{1 / (n+1)}-1\) and \(a_{n}=c^{1 / n }-1\). By multiplying these two inequalities, we get \(a_{n+1} + 1 = c^{1 / (n+1)} \leq c^{1 / n} = a_{n} + 1\), implying \(a_{n+1}\leq a_{n}\) and hence \(a_{n}\) is a decreasing sequence.

Step 3: Bounding the sequence

Now we need to prove that the sequence \(a_{n}\) is bounded below by 0 and above by \((c-1) /n\). To prove the upper bound, we use the binomial theorem for \((1+a_{n})^n\), and compare this with \(c = (1+a_n)^n \leq 1 + n a_n\). Rearranging gives \(a_n \leq (c-1)/n\). To prove the lower bound, we simply observe that \(a_n = c^{1/n} - 1\) is always greater than or equal to 0 for \(c>1\). Therefore, \(0 \leq a_{n} \leq(c-1) /n\).

Step 4: Applying the Monotone Convergence Theorem

Since \(a_n\) is decreasing and bounded, it must converge to a limit as per the Monotone Convergence Theorem. As \(a_n\) is bounded above by \((c-1)/n\), which tends to 0 as \(n\) tends to infinity, and it is also bounded below by 0, we conclude that the limit of \(a_n\) is 0.

Step 5: Proving the original assertion

As we have shown that \(\lim _{n \rightarrow \infty} a_n=0\), this implies \(\lim _{n \rightarrow \infty}(c^{1 / n} -1) = 0\), and hence that \(\lim _{n \rightarrow \infty} c^{1 / n}=1\).

Key Concepts

Monotone Convergence Theorem

The Monotone Convergence Theorem is a cornerstone of calculus concerning sequences. It states that if a sequence is both bounded and monotonic (either non-increasing or non-decreasing), it must converge to a limit. This theorem is incredibly useful because it allows us to establish the convergence of sequences without needing to find their exact limits.

In the context of our exercise, the sequence \(a_n = c^{1/n} - 1\) is shown to be decreasing, meaning that it satisfies the monotonicity condition. Additionally, by proving it is bounded (between 0 and \((c-1)/n\)), we ensure that it meets the criteria laid out by the Monotone Convergence Theorem. Hence, the sequence \(a_n\) converges to a limit.

Sequence Monotonicity

Monotonicity in a sequence is a property that describes how the sequence behaves as it progresses. A monotonic sequence either always increases or decreases, but never both.

For the sequence \(a_n = c^{1/n} - 1\), we determined it is decreasing. To prove this, you show that \(a_{n+1} \leq a_n\) for all \(n\). Simplifying the inequalities for our sequence shows this property holds, confirming monotonicity.

• Decreasing sequences: \(a_{n+1} \leq a_n\)
• Increasing sequences: \(a_{n+1} \geq a_n\)

This criterion is essential for applying certain theorems such as the Monotone Convergence Theorem, and it heavily influences the sequence's limits and bounds.

Upper and Lower Bounds in Sequences

In sequences, understanding upper and lower bounds helps ascertain if sequences stay within a certain range as they progress. An upper bound is a value which the sequence does not exceed, while a lower bound is a value it does not fall below.

For the sequence \(a_n = c^{1/n} - 1\):

• Upper Bound: Calculating or proving \(a_n \leq (c-1)/n\) proves that as \(n\) increases, this upper bound approaches zero, keeping the sequence in check.
• Lower Bound: As \(c^{1/n} - 1\) for \(c > 1\) is always positive, \(a_n \geq 0\) serves as the lower bound.

These bounds are crucial. They confirm that the sequence fits within defined limits, allowing us to apply the Monotone Convergence Theorem confidently.

Binomial Theorem in Series

The Binomial Theorem offers a method to expand expressions raised to a power, which comes in handy for bounding sequences in calculus.

For example, the exercise uses the binomial theorem to expand \((1 + a_n)^n\) and helps to compare it with \(c = (1 + a_n)^n \leq 1 + n a_n\). This inequality supports the conclusion that \(a_n \leq (c-1)/n\), crucial for setting the upper bound of the sequence. The binomial theorem thus becomes a powerful tool for simplifying and manipulating expressions, allowing for greater insight into otherwise complex problems.

• Useful for bounding arguments in sequences.
• Helps in proving inequalities crucial for understanding limits.

Employing this theorem makes complex series manageable, supporting broader calculus concepts like limits and convergence.