To analyze and solve the given mathematical expression, let's break it down step by step.

The expression is: $\sum_{i=2}^n \frac{1}{i^2} \leq 1 + \sum_{i=2}^n \frac{1}{i(i-1)}$

Step 1: Understanding the Left Side of the Inequality

The left side is a summation: $\sum_{i=2}^n \frac{1}{i^2}$ This represents the sum of the reciprocals of the squares of integers from $i=2$ to $n$.

Step 2: Analyzing the Right Side of the Inequality

The right side has two parts:

1. The constant 1.
2. The summation: $\sum_{i=2}^n \frac{1}{i(i-1)}$ Let's simplify this summation term.

Step 3: Simplifying the Summation on the Right Side

The term $\frac{1}{i(i-1)}$ can be rewritten using partial fractions: $\frac{1}{i(i-1)} = \frac{1}{i-1} - \frac{1}{i}$ So, the summation becomes: $\sum_{i=2}^n \left(\frac{1}{i-1} - \frac{1}{i}\right)$ This is a telescoping series, where most terms cancel out: $\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right)$ After cancellation, we are left with: $1 - \frac{1}{n}$

Step 4: Combining the Terms on the Right Side

So, the right side of the inequality simplifies to: $1 + \left(1 - \frac{1}{n}\right) = 2 - \frac{1}{n}$

Step 5: Forming the Final Inequality

The inequality now looks like: $\sum_{i=2}^n \frac{1}{i^2} \leq 2 - \frac{1}{n}$ This expresses that the sum of the reciprocals of squares from 2 to $n$ is less than or equal to $2 - \frac{1}{n}$.

Would you like further details or clarifications on any step?

Relative Questions:

1. How can telescoping series be applied to different types of sums?
2. What is the significance of inequalities in summations?
3. Can the sum $\sum_{i=2}^n \frac{1}{i^2}$ be bounded differently for large $n$?
4. How do partial fractions simplify complex rational expressions?
5. How does the value of $n$ affect the accuracy of the inequality?

Tip:

Understanding telescoping series is crucial for simplifying and evaluating complex summations. Look for patterns in the terms that cancel out in sequences.