How can I prove that the sequence $P_{n} = \sum_{j=0}^{n} \frac{1}{2^j}x^j$ is Cauchy? and converges to g.

Question

1- Why the author choose this type of polynomials to show that the inner product of all polynomials with the given inner product is not complete?

2-How can I prove that the sequence $P_{n} = \sum_{j=0}^{n} \frac{1}{2^j}x^j$ is Cauchy?

3-Why $P_{n} \rightarrow g$?

Could anyone explain this for me please?

Answer 1

1) $\sum_{j=0}^\infty (\frac{x}{2})^j$ is a Geometric series, which is known to converge when $|\frac{x}{2}| < 1$.

We are given that $x \in [0, 1]$, so $P_n$ converges. And since $P_n$ converges, it is Cauchy.

2) $g$ is the closed-form expression for the value of a Geometric series.

Answer 2

The series is alternating for $x<0$ so if it is Cauchy on some positive interval it is Cauchy on the corresponding negative interval. So assume $0\le x<2$. Show that the series is Cauchy for $x$.

Let $0<\epsilon<1$ and let $m>\dfrac{\ln\epsilon}{\ln\left(x/2\right)}$.

Then

\begin{eqnarray} m\ln\left(x/2\right)&<&\ln(\epsilon)\\ \left(\frac{x}{2}\right)^m&<&\epsilon \end{eqnarray}

Let $n>0$. Then

\begin{eqnarray} \left(\frac{x}{2}\right)^m-\left(\frac{x}{2}\right)^{m+n}&=&\left(\frac{x}{2}\right)^m\left(1-\left(\frac{x}{2}\right)^n\right)\\ &<&\left(\frac{x}{2}\right)^m\cdot1\\ &<&\epsilon \end{eqnarray}

Note: This is the answer I prepared for the original version of the question.