Exercises on Sequence Limits

Some worked examples involving sequence limits.

Exercise 1

In this exercise, we’re asked to prove that the limit of the sequence approaches 1.

$$ \lim_{n \rightarrow \infty} \frac{n}{2n+5} = 1 $$

According to the convergence theorem, a sequence converges to a real number $ l $ if:

$$ |a_n - l| < \epsilon $$

for every real number $\epsilon > 0$.

Since $a_n$ is defined for positive integers, we assume $n > 0$.

To verify the limit, we substitute $a_n$ with the expression of the sequence and set $l = 1$:

$$ \left|\frac{n}{2n+5} - 1\right| < \epsilon $$

We now simplify the expression inside the absolute value:

$$ \left|\frac{n - (2n+5)}{2n+5}\right| < \epsilon $$

$$ \left|\frac{-n - 5}{2n+5}\right| < \epsilon $$

Removing the minus sign in the numerator and dropping the absolute value:

$$ \frac{n + 5}{2n+5} < \epsilon $$

We proceed with further simplification:

$$ n + 5 < \epsilon(2n + 5) $$

$$ n + 5 < 2n\epsilon + 5\epsilon $$

$$ n - 2n\epsilon < 5\epsilon - 5 $$

$$ n(1 - 2\epsilon) < 5(\epsilon - 1) $$

This inequality can’t be simplified further. Now we check whether it holds for all values of $\epsilon > 0$.

If $\epsilon > 1$, the inequality holds because the left-hand side $n(1 - 2\epsilon)$ is negative, while the right-hand side $5(\epsilon - 1)$ is positive.
If $\epsilon = 1$, it still holds since $n(1 - 2\epsilon) = -n < 0$ for any $n > 0$.
If $\epsilon < 1$, the inequality does not always hold. In particular, for very small values of $\epsilon$, it fails. For example, when $\epsilon = 0.1$, the left-hand side becomes positive, while the right-hand side is negative.

Therefore, the sequence cannot converge to 1.

Let’s continue the analysis.

Note: The sequence is in fact convergent, but it converges to 1/2, not 1. So the actual limit as $n \to \infty$ is:

$$ \lim_{n \rightarrow \infty} \frac{n}{2n+5} = \frac{1}{2} $$

Let’s verify this using the same method:

$$ |a_n - l| < \epsilon $$

$$ \left|\frac{n}{2n+5} - \frac{1}{2} \right| < \epsilon $$

$$ \left|\frac{2n - (2n+5)}{2(2n+5)}\right| < \epsilon $$

$$ \left|\frac{-5}{2(2n+5)}\right| < \epsilon $$

$$ \frac{5}{2(2n+5)} < \epsilon $$

$$ \frac{5}{2\epsilon} < 2n + 5 $$

$$ \frac{5}{2\epsilon} - 5 < 2n $$

$$ \frac{5}{4\epsilon} - \frac{5}{2} < n $$

This inequality is satisfied for all values of $\epsilon > 0$, which confirms that the sequence converges to 1/2.