Problem 71 - Single Variable Calculus

Question

Let the sequence \(\left\\{a_{n}\right\\}\) be defined by $$ a_{n}=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}} $$ a. Show that \(\left\\{a_{n}\right\\}\) is increasing. b. Show that \(\left\\{a_{n}\right\\}\) is bounded above by establishing that \(a_{n}<2-1 / n\) for \(n \geq 2\) Hint: \(\frac{1}{n^{2}}<\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}\), for \(n \geq 2\) c. Using the results of parts (a) and (b), what can you deduce about the convergence of \(\left\\{a_{n}\right\\} ?\)

Step-by-Step Solution

Answer

In conclusion, we showed that the sequence \(\left\\{a_{n}\right\\}\) is increasing by comparing the terms \(a_n\) and \(a_{n+1}\), and we established that it is bounded above by using the hint provided and proving that \(a_n < 2 - \frac{1}{n}\) for \(n \geq 2\). By applying the Monotone Convergence Theorem, we deduce that the sequence \(\left\\{a_{n}\right\\}\) converges since it is both increasing and bounded above.

Step 1: Analyze the sequence definition

The given sequence \(\left\\{a_{n}\right\\}\) is defined as \(a_{n}=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}\). Analyzing the sequence definition shows that the terms are formed by adding a new fraction at each step, where the denominator is the square of the next integer in the sequence.

Step 2: Compare terms \(a_{n}\) and \(a_{n+1}\)

The term \(a_{n+1}\) will be formed by adding a new fraction \(\frac{1}{(n+1)^2}\) to \(a_n\). We can show that an extra positive fraction added to the previous term will make the sequence increasing: \[a_{n+1} = a_n + \frac{1}{(n+1)^2}\] Since \(\frac{1}{(n+1)^2}\) is always positive for all \(n \geq 1\), we have: \[a_{n+1} > a_n\] Thus, the sequence \(\left\\{a_{n}\right\\}\) is increasing. #b. Show that \(\left\\{a_{n}\right\\}\) is bounded above by establishing that \(a_{n}<2-1/n\) for \(n \geq 2\).#

Step 1: Utilize the hint

The hint given is \(\frac{1}{n^{2}}<\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}\), for \(n \geq 2\). We can substitute this inequality into the sequence definition to obtain: \[a_n < 1 + \sum_{k=2}^{n}\left(\frac{1}{k-1} - \frac{1}{k}\right)\]

Step 2: Simplify the sum

Now we can simplify the sum by breaking it down: \[\sum_{k=2}^{n}\left(\frac{1}{k-1} - \frac{1}{k}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right)\] After canceling out the common terms, we get: \[\sum_{k=2}^{n}\left(\frac{1}{k-1} - \frac{1}{k}\right) = 1 - \frac{1}{n}\]

Step 3: Substitute the simplified sum into the inequality

Now substitute this simplified sum back into the inequality from Step 1: \[a_n < 1 + (1 - \frac{1}{n})\] This simplifies to: \[a_n < 2 - \frac{1}{n}\] So the sequence \(\left\\{a_{n}\right\\}\) is bounded above by \(2 - \frac{1}{n}\) for \(n \geq 2\). #c. Using the results of parts (a) and (b), what can you deduce about the convergence of \(\left\\{a_{n}\right\\} ?\)

Step 6: Conclusion: Apply the Monotone Convergence Theorem

Since we have shown that \(\left\\{a_{n}\right\\}\) is increasing and bounded above, we can apply the Monotone Convergence Theorem. According to this theorem, an increasing and bounded above sequence is convergent. Therefore, we can conclude that the sequence \(\left\\{a_{n}\right\\}\) converges.

Key Concepts

Monotone Sequences

In mathematics, a monotone sequence is one that is either entirely non-increasing or non-decreasing. Specifically, if each term is greater than or equal to the previous term, the sequence is termed monotonically increasing. On the other hand, if each term is less than or equal to the preceding term, it's called monotonically decreasing.

For instance, the sequence given by \(a_{n}=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{n^{2}}\) is a classic example of a monotonic sequence. As we add a new fraction whose denominator is a squared natural number at each step, the sequence naturally grows because each of these fractions is positive. This makes the sequence increasingly larger with each successive term, confirming it as a monotonically increasing sequence.

Bounded Sequences

A sequence is said to be bounded if there exists some fixed number that all terms of the sequence are less than (bounded above) or greater than (bounded below). In simple terms, a bounded sequence is confined within a numerical limit and doesn't stretch to infinity. This concept is crucial when discussing convergence, as bounded sequences often have interesting properties related to their limits.

For the sequence from our exercise, \(a_{n}\), we demonstrated that it is bounded above by showing \(a_{n}<2-1/n\) for all \(n \geq 2\). This proves that no matter how large \(n\) gets, the terms of \(a_{n}\) will never exceed 2. This upper bound ensures that the sequence doesn’t veer off to infinity, keeping it within a certain range.

Series

A series in mathematics refers to the sum of the terms of a sequence. It's essentially what you get when you add up elements from a sequence indefinitely. Series can converge or diverge, meaning they can add up to a finite value or grow infinitely large. The behavior of a series, whether it tends to a limit or not, is central to the study of sequences and series.

The sequence \(a_{n}\) we are examining is actually a partial sum of the series formed by \(\frac{1}{n^2}\). When we talk about the convergence of \(a_{n}\), we're interested in what happens as \(n\) approaches infinity; does it approach a fixed number (converge) or not (diverge)? In this case, \(a_{n}\) represents the sum of terms up to \(n\), and as \(n\) grows, the additional terms become smaller, suggesting convergence.

Monotone Convergence Theorem

The Monotone Convergence Theorem is a fundamental principle in real analysis that gives us criteria for when a sequence will converge. According to this theorem, if a sequence is both monotone and bounded, then it must be convergent. In other words, if a sequence is increasing and has an upper bound, or if it is decreasing and has a lower bound, then it will always tend towards a limit.

Applying this theorem to our sequence \(a_{n}\), which we've determined is both increasing and bounded above, we can confidently conclude that it converges. This is a powerful notion because it tells us that under these conditions, the sequence \(a_{n}\) approaches a specific number as \(n\) becomes very large, allowing us to discuss the properties of this sequence's limit.