# Using the definition of convergence, how do you prove that the sequence {2^ -n} converges from n=1 to infinity?

Nov 30, 2016

Use the properties of the exponential function to determine N such as $| {2}^{- n} - {2}^{- m} | < \epsilon$ for every $m , n > N$

#### Explanation:

The definition of convergence states that the $\left\{{a}_{n}\right\}$ converges if:

$\forall \epsilon > 0 \text{ " EE N: AA m,n>N " } | {a}_{n} - {a}_{m} | < \epsilon$

So, given $\epsilon > 0$ take $N > {\log}_{2} \left(\frac{1}{\epsilon}\right)$ and $m , n > N$ with $m < n$

As $m < n$, $\left({2}^{- m} - {2}^{- n}\right) > 0$ so $| {2}^{- m} - {2}^{- n} | = {2}^{- m} - {2}^{- n}$

${2}^{- m} - {2}^{- n} = {2}^{- m} \left(1 - {2}^{m - n}\right)$

Now as ${2}^{x}$ is always positive, $\left(1 - {2}^{m - n}\right) < 1$, so

${2}^{- m} - {2}^{- n} < {2}^{- m}$

And as ${2}^{- x}$ is strictly decreasing and $m > N > {\log}_{2} \left(\frac{1}{\epsilon}\right)$

2^(-m) - 2^(-n) < 2^(-m) < 2^(-N) < 2^(-log_2(1/epsilon)

But:

${2}^{- {\log}_{2} \left(\frac{1}{\epsilon}\right)} = {2}^{{\log}_{2} \left(\epsilon\right)} = \epsilon$

So:

$| {2}^{- m} - {2}^{- n} | < \epsilon$

Q.E.D.