Question

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

Answer 1

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 `{a_n}` converges if:

`AA epsilon > 0 " " EE N: AA m,n>N " " |a_n-a_m| < epsilon`

So, given `epsilon >0` take `N > log_2(1/epsilon)` and `m,n > N` with `m < n`

As `m < n`, `(2^(-m) - 2^(-n) )> 0` so `|2^(-m) - 2^(-n)| = 2^(-m) - 2^(-n)`

`2^(-m) - 2^(-n) = 2^(-m)(1- 2^(m-n))`

Now as `2^x` is always positive, `(1- 2^(m-n) ) < 1`, so

`2^(-m) - 2^(-n) < 2^(-m)`

And as `2^(-x)` is strictly decreasing and `m > N > log_2(1/epsilon)`

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

But:

`2^(-log_2(1/epsilon) )= 2^(log_2(epsilon)) = epsilon`

So:

`|2^(-m) - 2^(-n)| < epsilon`

Q.E.D.