Question #2538f

1 Answer
Dec 27, 2017

My hunch is to use the Ratio Test to prove the #lArr# direction and the Divergence Test to prove the contrapositive of the #rArr# direction (which also proves the #rArr# implication). Not sure if it will work, but here goes.

Explanation:

(#lArr#) Suppose that #(u_{n})# is a sequence that converges to a negative number #u# and that #a>1#. Let #b_{n}=a^{u_{1}+u_{2}+cdots+u_{n}}=a^{u_{1}} * a^{u_{2}} cdots a^{u_{n}}#.

Then #|b_{n+1}|/|b_{n}|=(a^{u_{1}} * a^{u_{2}} cdots a^{u_{n}}*a^{u_{n+1}})/(a^{u_{1}} * a^{u_{2}} cdots a^{u_{n}})=a^{u_{n+1}}->a^{u}# as #n->infty#.

Since #a>1# and #u<0#, #|a^{u}|=a^{u}<1#.

Therefore, by the Ratio Test, the series #sum_{n=1}^{infty}(a^{u_{1}+u_{2}+cdots+u_{n}})# converges.

Therefore, the #lArr# direction is proven.

(#rArr#) Suppose that #(u_{n})# is a sequence that converges to a negative number #u# and that #a <= 1#. Choose #N in NN# so that #u_{n}< 0# for all #n > N#.

Let #A=a^{u_{1}+u_{2}+cdots+u_{N}}=a^{u_{1}} * a^{u_{2}} cdots a^{u_{N}}# and note that #A !=0#.

Next, suppose #n > N#. Then #a^{u_{1}+u_{2}+cdots u_{n}}=a^{u_{1}} * a^{u_{2}} cdots a^{u_{N}} * a^{u_{N+1}} cdots a^{u_{n}}=A * a^{u_{N+1}} cdots a^{u_{n}}#. Since #a <= 1# and #u_{N+1}<0#, #u_{N+2}<0#, ..., #u_{n}<0#, it follows that #a^{u_{N+1}} cdots a^{u_{n}}# is the product of a bunch of numbers that are all #>= 1#.

Since #A !=0#, it follows that #A * a^{u_{N+1}} cdots a^{u_{n}}# does not converge to zero as #n->infty#. But this means that #a^{u_{1}+u_{2}+cdots u_{n}}# does not converge to zero as #n -> infty#.

By the Divergence Test, the series #sum_{n=1}^{infty}(a^{u_{1}+u_{2}+cdots+u_{n}})# diverges.

Thus, the contrapositive of the #rArr# implication is proved, meaning the original implication itself is also proved.