Question

Find the errors and solve the problem correctly!
Thanks

Answer 1

The sequence converges to 5 as n goes to ∞. We can prove this rigorously:

The claim is that, for $n\in\Bbb N$,

$\displaystyle \lim_{n\to\infty} \frac{5n-1}{n+1} = 5$

By definition of the limit, this means for any ε > 0, there exists some N such that for any n larger than N, we have

$\left| \dfrac{5n-1}{n+1} - 5 \right| < \varepsilon$

In order to ensure this inequality, we work backwards to find N in terms of ε :

$\left| \dfrac{5n-1}{n+1} - 5 \right| = \left| \dfrac{5n-1 - 5(n+1)}{n+1} \right| = \dfrac4{|n+1|} < \varepsilon$

n is a natural number, so n + 1 > 0 and |n + 1| = n + 1. Then

$\dfrac4{n+1} < \varepsilon \implies \dfrac{n+1}4 > \dfrac1\varepsilon \implies n+1 > \dfrac4\varepsilon \implies n > \dfrac4\varepsilon - 1$

Let N be the smallest natural number greater than or equal to 4/ε - 1, i.e.

$N = \left\lceil \dfrac4\varepsilon - 1 \right\rceil$

Then the proof follows: suppose

$n \ge N = \left\lceil \dfrac4\varepsilon - 1 \right\rceil$

Then

$n \ge \left\lceil\dfrac4\varepsilon - 1\right\rceil \implies n > \dfrac4\varepsilon - 1 \\\\ \implies n+1 > \dfrac4\varepsilon \\\\ \implies \dfrac4{n+1} < \varepsilon \\\\\implies \left|\dfrac{5n-1}{n+1}-5\right| < \varepsilon$

as required.

Or we can avoid the rigor and use some facts we know about limits:

$\displaystyle \lim_{n\to\infty} \frac{5n-1}{n+1} = \lim_{n\to\infty} \frac{5-\frac1n}{1+\frac1n} = 5$

since 1/n converges to 0 as n goes to ∞.