Math Problem Statement

Resolver el exámen por favor

Resolver el exámen por favor

Solution

I can help solve the exam shown in the image. I'll extract and solve the problems step by step.

Let's begin by analyzing the content of the exam from the image.


Question 1:

Complete the following series:

n=1an\sum_{n=1}^{\infty} a_n

The specific details of the series and method to complete it are unclear. Could you please clarify or describe the series in question?


Question 2:

Determine if the following sequence converges or diverges:

2n23n3n2+5\frac{2n^2 - 3n}{3n^2 + 5}

I'll start solving this problem right away.

To determine if the sequence converges or diverges, we need to analyze the limit as nn \to \infty:

limn2n23n3n2+5\lim_{n \to \infty} \frac{2n^2 - 3n}{3n^2 + 5}

Dividing the numerator and the denominator by n2n^2, we get:

limn23n3+5n2\lim_{n \to \infty} \frac{2 - \frac{3}{n}}{3 + \frac{5}{n^2}}

As nn \to \infty, the terms 3n\frac{3}{n} and 5n2\frac{5}{n^2} approach 0, so the expression simplifies to:

23\frac{2}{3}

Thus, the sequence converges to 23\frac{2}{3}.


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Math Problem Analysis

Mathematical Concepts

Series
Sequence Convergence
Limits
Summations
Calculus

Formulas

\sum_{n=1}^{\infty} a_n
\lim_{n \to \infty} \frac{2n^2 - 3n}{3n^2 + 5}

Theorems

Limit of a Sequence
Convergence of a Sequence

Suitable Grade Level

University level