Problem 64 - Algebra and Trigonometry

Question

Write the sum using sigma notation. $$\frac{\sqrt{1}}{1^{2}}+\frac{\sqrt{2}}{2^{2}}+\frac{\sqrt{3}}{3^{2}}+\cdots+\frac{\sqrt{n}}{n^{2}}$$

Step-by-Step Solution

Answer

\( \sum_{k=1}^{n} \frac{\sqrt{k}}{k^2} \).

Step 1: Identify the General Term

Observe the pattern in each term of the sum. Each term of the sequence can be described as \( a_k = \frac{\sqrt{k}}{k^2} \). This general term will be essential for writing the sum in sigma notation.

Step 2: Determine the Range of the Index

Identify the range of the sequence, which starts at \( k=1 \) and ends at \( k=n \), as given in the sequence: \( \frac{\sqrt{1}}{1^2}, \frac{\sqrt{2}}{2^2}, \dots, \frac{\sqrt{n}}{n^2} \).

Step 3: Write the Sum in Sigma Notation

Combine the general term and the range of the index to express the sum using sigma notation. The expression becomes:\[ \sum_{k=1}^{n} \frac{\sqrt{k}}{k^2} \].This notation captures the entire sequence from \( k=1 \) to \( k=n \).

Key Concepts

Sequence

A sequence is an ordered list of numbers that follows a specific pattern or rule. In mathematics, sequences can be finite or infinite, and they play a crucial role in various areas of math, such as calculus and algebra.
Understanding sequences is essential when dealing with sigma notation, as you often need to identify the pattern to express a series as a sum.Sequences are defined by their terms. Consider the sequence in the original exercise: \[ \left\{ \frac{\sqrt{1}}{1^2}, \frac{\sqrt{2}}{2^2}, \frac{\sqrt{3}}{3^2}, \ldots, \frac{\sqrt{n}}{n^2} \right\} \]This sequence starts from one and increases by one for each subsequent term. Each term is defined by the expression \( \frac{\sqrt{k}}{k^2} \).
It's a sequence because each term's structure follows a regular pattern as defined by this formula.

• Finite Sequences: These have a definite number of terms; in the example, the sequence ends at \( n \).
• Infinite Sequences: Continue indefinitely with no ending term.

Recognizing the pattern of sequences helps in formulating the general term, which is pivotal for sigma notation.

General Term

The general term of a sequence expresses each term in the sequence using a variable. This term is crucial because it encapsulates the pattern or rule that defines the entire sequence.
When you find the general term, you can represent all terms in a sequence without listing them individually.In the original exercise, the general term for the sequence is identified as:\[ a_k = \frac{\sqrt{k}}{k^2} \]This formula can plug in any integer \( k \) to generate the corresponding term of the sequence, making the sequence representation succinct and elegant. Here's how the general term works:

• Substitute the index: Plug \( k = 1 \) into the formula to get \( \frac{\sqrt{1}}{1^2} = 1 \).
• Continue substituting: For \( k = 2 \), the term becomes \( \frac{\sqrt{2}}{4} \).
• Try the last term: For \( k = n \), obtain the term \( \frac{\sqrt{n}}{n^2} \).

The general term is pivotal in writing sequences in a compact manner using sigma notation.

Index Range

The index range specifies where the sequence begins and ends. This range is crucial for constructing the sigma notation expression, as it defines which terms are included in the sum.
It essentially provides the boundaries for the sequence in question.In our original problem, the sequence starts at \( k = 1 \) and ends at \( k = n \). This implies that the series includes all terms starting from the first index and concludes at the \( n^{th} \) term.For the given sequence in the exercise:

• The starting point is \( k = 1 \).
• The ending point is \( k = n \).

These limits are critical when using sigma notation as they tell you exactly where to start and stop the summation.
In sigma notation, this range is written under and above the summation symbol \( \sum \). The complete expression as derived in the exercise is:\[ \sum_{k=1}^{n} \frac{\sqrt{k}}{k^2} \]Understanding the index range ensures that your summation is accurate and complete, including all necessary terms for the pattern presented.