Question

Suppose that a sequence of numbers \(a_{n}>0\) has the property that \(a_{1}=1\) and \(a_{n+1}=\frac{1}{(n+1)^{2}} S_{n}\), where \(S_{n}=a_{1}+\cdots+a_{n} .\) Can you determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges? (Hint. \(S_{2}=a_{2}+a_{1}=a_{2}+S_{1}=a_{2}+1=1+1 / 4=(1+1 / 4) S_{1}\) \(S_{3}=\frac{1}{3^{2}} S_{2}+S_{2}=(1+1 / 9) S_{2}=(1+1 / 9)(1+1 / 4) S_{1}\), etc. Look at \(\ln \left(S_{n}\right)\), and use \(\left.\ln (1+t) \leq t, t>0 .\right)\)

Short answer

The series \(\sum_{n=1}^{\infty} a_n\) converges.

Step-by-step solution

Understanding the Problem

We have a sequence defined by initial terms and a recursive formula. We need to find whether the series \(\sum_{n=1}^{\infty} a_n\) converges. Given \(a_1 = 1\), \(a_{n+1} = \frac{1}{(n+1)^2} S_n\), where \(S_n = a_1 + \cdots + a_n\). We will use given hints and properties of logs.

Compute Initial Terms

Calculate initial terms to see how they behave. Start with \(a_1 = 1\). By iterating, \(a_2 = \frac{1}{4}S_1 = \frac{1}{4}\times 1 = \frac{1}{4}\). Then \(\ S_2 = a_1 + a_2 = 1 + \frac{1}{4} = \frac{5}{4}\).

Recursive Strategy

From the recursive formula, \(a_{n+1} = \frac{1}{(n+1)^2} S_n\), we see that each \(a_{n+1}\) is defined by all previous terms. This is crucial for finding \(S_n\) recursively.

Exploit the Given Hint

Using the hint, \(S_2 = (1+\frac{1}{4})S_1\) and \(S_3 = (1 + \frac{1}{9})S_2\), we find a pattern. It yields \(S_n = (1+\frac{1}{4})(1+\frac{1}{9})\cdots (1+\frac{1}{n^2})S_1\). Realize \(\ln(S_n) = \ln((1+\frac{1}{4})(1+\frac{1}{9})\cdots(1+\frac{1}{n^2})S_1)\).

Apply Logarithm Bound

Logarithm inequality \(\ln(1+t) \leq t\) helps. \(\ln(S_n) \leq \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2}\). Sum \(\frac{1}{k^2}\) for \(k\), is convergent (since \(\sum 1/k^{2}\) converges, a known result).

Conclude Convergence

Since the upper bound \(\sum_{k=2}^n \frac{1}{k^2}\) converges, \(\ln(S_n)\) is bounded, meaning \(S_n\) is bounded. Therefore, \(\sum_{n=1}^{\infty} a_n\) converges by comparing to the bounded growth allowed by the log inequality.

Key concepts

Recursive Sequences

Recursive sequences are types of sequences where each term is defined based on one or more of the preceding terms. They follow a specific recursive formula, and can often model many real-world processes. For the sequence in this exercise, the recursive formula is given by \(a_{n+1} = \frac{1}{(n+1)^2} S_n\). This means each new term \(a_{n+1}\) is calculated using the sum of all previous terms \(S_n\). This recursive formula allows for building the sequence step by step, starting from known initial conditions such as \(a_1 = 1\). This is crucial as it enables us to predict the behavior of the sequence at very large indices by understanding how each term is formed based on its predecessors.

Logarithm Properties

Logarithm properties are powerful tools in mathematics, which simplify complex multiplicative relationships into additive ones. In this exercise, a particular property of the natural logarithm is used: \(\ln(1+t) \leq t\), for \(t > 0\). This represents the idea that the logarithm of a number slightly more than one is approximately equal to that slightly extra amount, \(t\). Applying this to the sequence, we can transform the multiplicative sequence of \(S_n\) into an additive form, \(\ln(S_n)\), making it easier to handle and bound. The process helps in evaluating the behavior of the sequence components individually and is instrumental in proving the convergence of \(\sum_{n=1}^{\infty} a_{n}\).

Series Convergence

Series convergence is a fundamental concept in calculus concerning whether the sum of the terms in a series approaches a finite limit as the number of terms increases. In simpler terms, it checks if, when you add up an infinite number of terms, you get a number that is not infinite. For the series \(\sum_{n=1}^{\infty} a_n\) in this problem, determining convergence involves checking if the summed sequence of terms reaches a bounded value. We use the comparison test, among other techniques, where we identify that since the sum \(\sum \frac{1}{k^2}\) is convergent, any series bounded by this known convergent series is also convergent. This comparison leverages known results, simplifying the demonstration of convergence without having to calculate each term individually.

Bounded Sequences

Bounded sequences are sequences whose terms stay within a fixed interval no matter how many terms you consider. This feature is significant because it allows us to use particular mathematical tools and theorems to draw conclusions about the behavior and sum of the sequence. In the context of this exercise, the concept of bounded sequences ties into the convergence of \(S_n\). Since \(\ln(S_n)\) is shown to be bounded due to the convergence of \(\sum \frac{1}{k^2}\), the sequence \(S_n\) itself must also be bounded. This also implies that the terms \(a_n\) become sufficiently small so that their infinite sum \(\sum_{n=1}^\infty a_n\) converges. Bounded sequences therefore play a crucial role in ensuring that the sequence does not "blow up" to infinity, affirming the convergence of the series.