Definition: Given a sequence < a(n) > we say that it converges if there exists A

Question

Definition: Given a sequence < a(n) > we say that it converges if there exists A in element R such that for all b > 0, there exists N in element                    N such that for all n > N we have | A - a(n) | < b.

Using no negative words say what it would mean for a sequence < a(n) > not to converge. Then, show that < 3^(2N - 1) > does not converge.

Hint:

(A.) Take an arbitrary M in element R.

(B.) Explain why 3^(2N - 1) >= M if and only if (2N - 1)log[10] (3) > log[10] (M).

       (NOTE: log[10] (3) means log base 10 (3))

(C.) Use this to get an express of the form 3^(2N - 1) >= M if and only if N >= (an expression involving M and logs).

(D.) Now choose N accordingly.

Solve parts A thru D and show work for full points.

Explanation / Answer

For <a(n)> not to converge

For evry A that belongs to R and every b>0 there exists an N in element N such that for all n > N we have | A - a(n) | >= b i,e the term | A - a(n) | does not tend to 0

Taking any M in R, M>0 such that

3^(2n - 1) >= M

so taking log on both sides

log[10] (3^(2n-1)) >= log[10] (M). so (2n - 1)log[10] (3) > log[10] (M)..............sufficient condition

similarly

(2n - 1)log[10] (3) > log[10] (M) implies log[10] (3^(2n-1)) >= log[10] (M).

Taking e to the power both sides 3^(2n - 1) >= M.....................................necessary condition

so

3^(2n - 1) >= M if and only if (2n - 1)log[10] (3) >= log[10] (M).

so 2nlog[10] (3) >= log[10] (M)+log[10] (3)

so 2nlog[10] (3) >= log[10] (3M)

so n>=1/2 * log[3](3M)

so n>=1/2 + log[3](sqrt(M))

so n>=ceil( 1/2 + log[3](sqrt(M)) )

So for every choice of M in R and every b>0 we get ceil( 1/2 + log[3](sqrt(M)) )such that for all ceil(n>1/2 + log[3](sqrt(M)) )

the term |M-3^(2n - 1)|>=b i.e the term does not tend to zero......................does not converge

Note- ceil(x) is smallest integer greater than or equal to x

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