We are tasked to find the limit of the sequence $c_n$ given the recurrence relation:

$c_{n+1} = \left(1 - \frac{1}{n}\right)c_n + b_n,$ where $\lim_{n \to \infty} b_n n^2 = 0$.

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Step 1: Observations

1. The factor $\left(1 - \frac{1}{n}\right)$ in the recurrence relation suggests that the sequence $c_n$ may be governed by a weighted sum of its past values, modified slightly by $b_n$.
2. The condition $\lim_{n \to \infty} b_n n^2 = 0$ implies that $b_n$ approaches 0 faster than $\frac{1}{n^2}$ as $n \to \infty$.

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Step 2: Rewrite the Recurrence Relation

Define $c_n = d_n + \tilde{c}_n$, where $d_n$ solves the homogeneous part:

$d_{n+1} = \left(1 - \frac{1}{n}\right)d_n.$

The solution to the homogeneous equation can be written as:

$d_n = c_1 \prod_{k=1}^{n-1} \left(1 - \frac{1}{k}\right).$

Using the property $\prod_{k=1}^{n-1} \left(1 - \frac{1}{k}\right) = \frac{1}{n}$, we get:

$d_n = \frac{c_1}{n}.$

Thus, $d_n \to 0$ as $n \to \infty$.

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Step 3: Particular Solution and Total Sequence

Now consider the particular solution $\tilde{c}_n$ due to $b_n$. Assume that as $n \to \infty$, $\tilde{c}_n \to L$, a constant. Substituting into the recurrence relation:

$L = \left(1 - \frac{1}{n}\right)L + b_n.$

For large $n$, this simplifies to $b_n \to 0$ due to $\lim_{n \to \infty} b_n n^2 = 0$. Hence, the total solution $c_n$ asymptotically approaches 0.

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Final Answer

$\lim_{n \to \infty} c_n = 0.$

Would you like a more detailed explanation or derivation of any specific step?

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Relative Questions

1. What happens if $\lim_{n \to \infty} b_n n^2 \neq 0$?
2. How does the solution change if $b_n = \frac{1}{n^3}$?
3. Can this method be applied to more general recurrence relations of this form?
4. How is the homogeneous solution $d_n$ related to the series sum?
5. What are the conditions on $b_n$ for $c_n$ to converge?

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Tip:

Always analyze the growth rate of terms in recurrence relations to identify dominant behavior over large $n$.