lim(((1+3+...+(2*n-1))/(n+1))-((2*n+1)/(2)))=

We are tasked with evaluating the limit:

\[
\lim_{n \to \infty} \left( \frac{1 + 3 + 5 + \cdots + (2n - 1)}{n + 1} - \frac{2n + 1}{2} \right)
\]

### Step 1: Simplifying the Sum of Odd Numbers
The sum of the first \( n \) odd numbers \( 1 + 3 + 5 + \cdots + (2n - 1) \) is a well-known result:

\[
1 + 3 + 5 + \cdots + (2n - 1) = n^2
\]

Thus, we can replace the sum in the expression:

\[
\lim_{n \to \infty} \left( \frac{n^2}{n + 1} - \frac{2n + 1}{2} \right)
\]

### Step 2: Simplifying Each Term
Now, simplify the two terms separately:

1. \(\frac{n^2}{n+1}\):

\[
\frac{n^2}{n+1} = \frac{n^2}{n(1 + \frac{1}{n})} = \frac{n}{1 + \frac{1}{n}} = \frac{n}{1 + \frac{1}{n}} \to n \text{ as } n \to \infty
\]

2. \(\frac{2n + 1}{2}\):

\[
\frac{2n + 1}{2} = n + \frac{1}{2}
\]

### Step 3: Subtract the Two Expressions
Now subtract the two simplified terms:

\[
\lim_{n \to \infty} \left( n - \left( n + \frac{1}{2} \right) \right)
\]

This simplifies to:

\[
\lim_{n \to \infty} \left( n - n - \frac{1}{2} \right) = \lim_{n \to \infty} -\frac{1}{2} = -\frac{1}{2}
\]

### Final Answer:
\[
\boxed{-\frac{1}{2}}
\]

Would you like further clarification on any steps? Here are some related questions:

1. How can we generalize sums of odd numbers?
2. What is the limit of rational functions as \( n \to \infty \)?
3. How do we apply the squeeze theorem to evaluate limits?
4. What other series converge in a similar manner?
5. What are other common sums involving consecutive odd or even numbers?

**Tip**: When dealing with limits at infinity, try to simplify each term by factoring out dominant terms for easier analysis.

Explore how to evaluate the limit lim(((1+3+...+(2*n-1))/(n+1))-((2*n+1)/(2))) as n approaches infinity. This problem demonstrates key concepts such as the sum of odd numbers, limits, and simplification of rational expressions. Suitable for undergraduate calculus students.

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