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

To prove that the sequence \( \{2^{-n}\} \) converges as \( n \) approaches infinity, we use the definition of convergence, which states that a sequence \( \{a_n\} \) converges to a limit \( L \) if, for every positive real number \( \epsilon \), there exists a positive integer \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| < \epsilon \). In the case of the sequence \( \{2^{-n}\} \), we aim to show that it converges to 0 as \( n \) approaches infinity. Let \( \epsilon > 0 \) be given. We need to find a positive integer \( N \) such that for all \( n > N \), \( |2^{-n} - 0| < \epsilon \). Since \( 2^{-n} \) approaches 0 as \( n \) becomes larger, we can choose \( N \) such that \( 2^{-N} < \epsilon \). This ensures that for all \( n > N \), \( |2^{-n} - 0| = 2^{-n} < 2^{-N} < \epsilon \), which satisfies the definition of convergence. Therefore, by choosing \( N \) appropriately based on the given \( \epsilon \), we have shown that the sequence \( \{2^{-n}\} \) converges to 0 as \( n \) approaches infinity.

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)#

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

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