Math Problem Statement

prove an = n/(n^2+1) + n/(n^2+2) .... n/(n^2+n) is a convergent sequence using sandwich theorem

Solution

To prove that the sequence

an=k=1nnn2+ka_n = \sum_{k=1}^{n} \frac{n}{n^2 + k}

is convergent using the sandwich (squeeze) theorem, we need to find two sequences, say bnb_n and cnc_n, such that:

bnancnfor all nb_n \leq a_n \leq c_n \quad \text{for all } n

and both bnb_n and cnc_n converge to the same limit as nn \to \infty. Here's how we proceed:

Step 1: Express the General Term

The given sequence ana_n is a sum of terms of the form

nn2+k,k=1,2,,n\frac{n}{n^2 + k}, \quad k = 1, 2, \dots, n

Let's analyze the behavior of each individual term. For large nn, we approximate:

nn2+knn2=1n\frac{n}{n^2 + k} \approx \frac{n}{n^2} = \frac{1}{n}

Thus, for large nn, each term behaves like 1n\frac{1}{n}.

Step 2: Find Upper Bound cnc_n

Since k1k \geq 1, we have:

n2+kn2+1n^2 + k \geq n^2 + 1

Thus,

nn2+knn2+1\frac{n}{n^2 + k} \leq \frac{n}{n^2 + 1}

Summing this inequality over all kk from 1 to nn, we get:

an=k=1nnn2+kk=1nnn2+1=nnn2+1a_n = \sum_{k=1}^{n} \frac{n}{n^2 + k} \leq \sum_{k=1}^{n} \frac{n}{n^2 + 1} = n \cdot \frac{n}{n^2 + 1}

Simplifying the right-hand side:

ann2n2+1a_n \leq \frac{n^2}{n^2 + 1}

For large nn, n2n2+11\frac{n^2}{n^2 + 1} \to 1.

Thus, we have an upper bound:

an1a_n \leq 1

Step 3: Find Lower Bound bnb_n

Similarly, since knk \leq n, we have:

n2+kn2+nn^2 + k \leq n^2 + n

Thus,

nn2+knn2+n=nn(n+1)=1n+1\frac{n}{n^2 + k} \geq \frac{n}{n^2 + n} = \frac{n}{n(n + 1)} = \frac{1}{n + 1}

Summing this inequality over all kk from 1 to nn, we get:

an=k=1nnn2+kk=1n1n+1=n1n+1a_n = \sum_{k=1}^{n} \frac{n}{n^2 + k} \geq \sum_{k=1}^{n} \frac{1}{n + 1} = n \cdot \frac{1}{n + 1}

Simplifying the right-hand side:

annn+1a_n \geq \frac{n}{n + 1}

For large nn, nn+11\frac{n}{n + 1} \to 1.

Thus, we have a lower bound:

an11n+1a_n \geq 1 - \frac{1}{n + 1}

Step 4: Apply the Sandwich (Squeeze) Theorem

Now, we have the inequalities:

11n+1ann2n2+11 - \frac{1}{n + 1} \leq a_n \leq \frac{n^2}{n^2 + 1}

As nn \to \infty:

11n+11andn2n2+111 - \frac{1}{n + 1} \to 1 \quad \text{and} \quad \frac{n^2}{n^2 + 1} \to 1

By the sandwich theorem, we conclude that:

an1asna_n \to 1 \quad \text{as} \quad n \to \infty

Thus, the sequence ana_n is convergent, and its limit is 11.


Would you like more details or have any questions? Here are 5 related questions to expand your understanding:

  1. How does the behavior of the general term nn2+k\frac{n}{n^2 + k} affect the convergence?
  2. Can the sandwich theorem be applied to sequences involving sums in more complex cases?
  3. How do we find the upper and lower bounds in different convergence problems?
  4. Why is it important that both bounds converge to the same limit in the sandwich theorem?
  5. How would this proof change if the terms were nn2+k2\frac{n}{n^2 + k^2}?

Tip: When using the sandwich theorem, the key is finding bounds that closely approximate the behavior of the sequence for large values of nn.

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Math Problem Analysis

Mathematical Concepts

Sequences and Series
Convergence
Approximation

Formulas

a_n = \sum_{k=1}^{n} \frac{n}{n^2 + k}
Upper Bound: a_n \leq \frac{n^2}{n^2 + 1}
Lower Bound: a_n \geq \frac{n}{n + 1}

Theorems

Sandwich Theorem (Squeeze Theorem)

Suitable Grade Level

University Level (Calculus I or II)