
Convergent Sequence
Hi, I think I am close to the answer for this but I am missing something:
Suppose $\displaystyle x_{n}$ is a sequence satisfying $\displaystyle x_{n+1}x_{n}<\frac{1}{2^n}$ for all n. Show that $\displaystyle x_{n}$ is a convergent sequence.
So to prove it is convergent, I need to prove it is a Cauchy sequence, ie for all $\displaystyle \epsilon$ there is an N such that $\displaystyle x_{m}x_{n}<\epsilon$ for all $\displaystyle n,m>N$.
Now $\displaystyle x_{m}x_{n}= x_{m}x_{m1}+x_{m1}...+x_{n+1}x_{n}$ and the triangle inequality gives $\displaystyle x_{m}x_{n}\leqx_{m}x_{m1}+x_{m1}x_{m2}+...+x_{n+1}x_{n}$.
Then using the inequality given in the question I get $\displaystyle x_{m}x_{n}<\frac{1}{2^{m1}}+\frac{1}{2^{m2}}+...+\frac{1}{2^n}<\frac{1}{2^n}+\frac{1}{2^n}+ ...+\frac{1}{2^n}=\frac{mn}{2^n}$.
This is where I get stuck, I think I need to get rid of that mn somehow.

Yes, you are quite close to the answer, right up to the inequality $\displaystyle x_{m}x_{n}< \frac{1}{2^{m1}}+\frac{1}{2^{m2}}+...+\frac{1}{2^n}$ (where m>n). Here, instead of estimating each individual term (replacing each term by the largest term $\displaystyle 1/2^n$), write the terms in the reverse order $\displaystyle \frac{1}{2^n} + \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}} + \ldots + \frac{1}{2^{m1}}$. This is a (finite) geometric series, whose sum is less than the infinite sum $\displaystyle \sum_{r=n}^\infty \frac1{2^r} = \frac1{2^{n1}}$.

Ah cheers, I never think of using geometric series!