Series Conversion, Jacobsthal Sequence

Question

I can't wrap my head around this exercise:

Show that

$$\sum_{i=0}^\infty \left(- \frac{1}{2}\right)^i = \frac{2}{3}$$

I started by calculating elements of the series:

$$k = 0 \rightarrow \frac{1}{1},\ k = 1 \rightarrow \frac{1}{2},\ k = 2 \rightarrow \frac{3}{4}, \rightarrow k = 3 \rightarrow \frac{5}{8},\ k = 4 \rightarrow \frac{11}{16}$$

and arrived at the conclusion that the formula of the partial sum is the following:

$$S_n = \frac{\text{Jacobsthal sequence}}{2^n} = \frac{a(n) = a(n - 1) + 2a(n - 2)\text{ with }a(0) = 0, a(1) = 1}{2^n} = \frac{\frac{2^n}{3}}{2^n} = \frac{1}{3} ?$$

The approximation used to describe the Jacobsthal sequence, namely to two to the n over 3, is not precise enough, thus I try to use induction to see if the partial sum holds, and then calculate the limit, but here is where I get stuck:

$$S_{n+1} = S_n + \left(-\frac{1}{2}\right)^n = \frac{a(n - 1) + 2a(n - 2)}{2^n} + \left(-\frac{1}{2}\right)^n =\ ...\ ?= \frac{a(n + 1)}{2^{n +1}}$$

any hints if this is on the right track and how can I manipulate the Jacobsthal series?

Thank you.

Answer 1

Here's another approach. We have:

$$ s_n = \sum_{i = 0}^{\infty} \left( - \frac{1}{2} \right)^i $$ $$ = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - ... $$

Which we can restate as:

$$ s_n = \sum_{i = 0}^{\infty} \frac{1}{4^i} - \frac{1}{2} \sum_{i = 0}^{\infty} \frac{1}{4^i} $$ $$ = \frac{1}{2} \sum_{i = 0}^{\infty} \frac{1}{4^i}$$

Computing the geometric series we get:

$$ s_n = \frac{1}{2} \frac{1}{1 - \frac{1}{4}} = \frac{2}{3}$$

Answer 2

Hint:

Define $$S_n=\sum_{i=0}^n\left(-\frac{1}{2}\right)^i$$

Then compare $S_n$ and $-\frac{1}{2}S_n$ to find a simpler expression for $S_n$.

You should then find that:

$$\left(1+\frac{1}{2}\right)S_n=1-\left(-\frac{1}{2}\right)^{n+1}$$ What can you then say about $\lim_{n\rightarrow\infty} S_n$?

Answer 3

$S_\infty ={a \over1-r} $

$\sum^\infty_{k=0} (-{1 \over 2} )^k = {1 \over 1+ {1\over 2}} = 1*{2\over3} = {2\over3}$