Hi everyone,
Can you help me to prove that
(1-b^(m+1))(-(b+1))/b^2<(1+b+b^2+b^3+...+b^m)
in logestic way
If you can try to answer me today because I have exam tomorrow in it
Thanks for any helping
Do you mean, for $\displaystyle b\not =0$:
$\displaystyle \frac{(1-b^{m+1})(-(b+1))}{b^2} < 1+b+b^2+...b^m$
Note that if $\displaystyle b=1$ then it is trivial.
Assume that $\displaystyle b\not = 1$ then:
$\displaystyle \frac{(1-b^{m+1})(-(b+1))}{b^2} < \frac{1-b^{m+1}}{1-b}$
If $\displaystyle |b|<1$ then $\displaystyle 1-b^{m+1}>0$ and hence:
$\displaystyle \frac{-(b+1)}{b^2} < \frac{1}{1-b}$
Which is true because LHS is negative and RHS is positive.
If $\displaystyle |b|>1$ then $\displaystyle 1-b^{m+1}<0$ and hence*:
$\displaystyle \frac{-(b+1)}{b^2} > \frac{1}{1-b}$
Thus,
$\displaystyle (b+1)(b-1) < b^2$
Thus,
$\displaystyle b^2 - 1 < b^2$
Which is true.
*)It also depends when $\displaystyle m$ is even or odd, but those a minor details.
This is the Geometric Sum.