Question

Let \(a_{n}=2^{-\mid n / 2]}\) where \([x]\) is the greatest integer less than or equal to \(x\). Determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges and justify your answer.

Short answer

The series \(\sum_{n=1}^{\infty} a_{n}\) converges to 4.

Step-by-step solution

Analyze the formula for \(a_n\)

Given the sequence \(a_{n} = 2^{-\lfloor n/2 \rfloor}\), this formula means that as \(n\) increases, \(n/2\) is taken, and the greatest integer less than or equal to \(n/2\) is used as the exponent with \(-1\) power of 2. This results in a sequence where each value is repeated twice: \(2^0, 2^0, 2^{-1}, 2^{-1}, 2^{-2}, 2^{-2}, \ldots\).

Express the sequence \(a_n\) explicitly

Let's list the terms explicitly based on the expression: \(a_1 = 2^0 = 1\), \(a_2 = 2^0 = 1\), \(a_3 = 2^{-1} = \frac{1}{2}\), \(a_4 = 2^{-1} = \frac{1}{2}\), \(a_5 = 2^{-2} = \frac{1}{4}\), \(a_6 = 2^{-2} = \frac{1}{4}\), and so on. This pattern continues as each term \(2^{-k}\) appears twice for each integer \(k\).

Construct a new series based on the repeated terms

We observe that the sequence is a patterned geometric series with each term repeated twice: \(1, 1, \frac{1}{2}, \frac{1}{2}, \frac{1}{4}, \frac{1}{4}, \ldots\). We can re-group them as \((1 + 1) + (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{4} + \frac{1}{4}) + \ldots\), simplifying to \(2 + 1 + \frac{1}{2} + \ldots\). This series is geometric.

Determine the convergence with geometric series test

Notice that the sequence \(2, 1, \frac{1}{2}, \ldots\) forms a geometric series with the first term \(a = 2\) and a common ratio \(r = \frac{1}{2}\). The series \(\sum_{k=0}^{\infty} 2 \cdot \left(\frac{1}{2}\right)^k\) is convergent because the common ratio \(r\) is \(\mid r \mid < 1\). The sum of this geometric series is \(\frac{a}{1-r} = \frac{2}{1-\frac{1}{2}} = 4\).

Key concepts

Convergence of Series

In mathematics, a series is a summation of terms of a sequence. Understanding when a series converges, or "comes together" to sum up to a particular value, is essential.
When analyzing a series, we check if the total of the terms approaches a specific limit as we add more and more terms. If it does, we say the series converges. If it doesn't and, instead, extensions are boundless or oscillate indefinitely, we say the series diverges.

• A basic rule is that a series \(\sum_{n=1}^{\infty} a_{n} \) converges if the sequence of partial sums \(s_{n} = a_{1} + a_{2} + \cdots + a_{n} \) approaches a finite limit as \(n \ ightarrow \infty\).
• An actionable test for convergence is to check if terms of the series get closer to zero as \(n\rightarrow \infty\). If they don't, then the series cannot converge.
• Geometric series, like the one in the given exercise, have a distinct pattern with a consistent ratio between terms. They converge if the absolute value of this common ratio \(r\) is less than one.

Geometric Progression

A geometric progression (or geometric sequence) is a series of numbers where each term after the first is found by multiplying the previous one by a constant, called the "common ratio."
In the exercise, we see a pattern where each term is repeated twice and forms the sequence, \(1, 1, \frac{1}{2}, \frac{1}{2}, \frac{1}{4}, \frac{1}{4}, \ldots\). This pattern can be described using the formula \[a_{n} = ar^{n-1}\]where \(a\) is the first term and \(r\) is the common ratio.

• For example, in this pattern: \(a = 2\) and \(r = \frac{1}{2}\).
• Geometric series can either converge or diverge based on the value of \(r\). If \(\mid r \mid < 1\), the series converges.
• The exercise's series is convergent because it satisfies this condition, and its sum can be calculated using the formula: \(\frac{a}{1 - r}\).

Understanding geometric progressions helps in recognizing patterns and solving convergence problems in series.

Greatest Integer Function

The greatest integer function, denoted by \[\lfloor x \rfloor\], refers to the greatest integer less than or equal to a given real number \(x\). This function is also commonly referred to as the floor function.
In simpler terms, this function 'rounds down' the given number to the nearest whole number.

• For example, \[\lfloor 3.7 \rfloor = 3\] and \[\lfloor -1.2 \rfloor = -2\].
• In the context of our exercise, it helps define the sequence \(a_{n} = 2^{-\lfloor n/2 \rfloor}\), thus controlling how the terms of the series change as \(n\) increases.
• The floor function is pivotal for forming patterns in sequences like the one given. It ensures the exponents change at calculated intervals.

This understanding allows us to predict and compute patterns or series that utilize integer-based modifications.