Using the definition of convergence, how do you prove that the sequence {5+(1/n)} converges from n=1 to infinity?

Jun 29, 2018

Let:

${a}_{n} = 5 + \frac{1}{n}$

then for any $m , n \in \mathbb{N}$ with $n > m$:

$\left\mid {a}_{m} - {a}_{n} \right\mid = \left\mid \left(5 + \frac{1}{m}\right) - \left(5 + \frac{1}{n}\right) \right\mid$

$\left\mid {a}_{m} - {a}_{n} \right\mid = \left\mid 5 + \frac{1}{m} - 5 - \frac{1}{n} \right\mid$

$\left\mid {a}_{m} - {a}_{n} \right\mid = \left\mid \frac{1}{m} - \frac{1}{n} \right\mid$

as $n > m \implies \frac{1}{n} < \frac{1}{m}$:

$\left\mid {a}_{m} - {a}_{n} \right\mid = \frac{1}{m} - \frac{1}{n}$

and as $\frac{1}{n} > 0$:

$\left\mid {a}_{m} - {a}_{n} \right\mid < \frac{1}{m}$.

Given any real number $\epsilon > 0$, choose then an integer $N > \frac{1}{\epsilon}$.

For any integers $m , n > N$ we have:

$\left\mid {a}_{m} - {a}_{n} \right\mid < \frac{1}{N}$

$\left\mid {a}_{m} - {a}_{n} \right\mid < \epsilon$

which proves Cauchy's condition for the convergence of a sequence.