Summation of power inequalities

Question

some idea for solve (?):

$x\in \mathbb{R}$

$\sum_{n=2}^{\infty } x^{n}\le 6$

I am at a loss :/

Thx

Answer 1

Let's pretend that our sum starts from $n=0$. Then we have $S=1+x+x^2+\ldots+x^{n-1}+x^n$. Let us then notice how $(1-x)S=1-x^{n+1}\iff S=\dfrac{1-x^{n+1}}{1-x\quad\ }$ . Now, as n tends towards $\infty$, if x were greater than $1$ in absolute value, then $S\to\pm\infty$. So $|x|<1$, in which case, $S=\dfrac1{1-x}$ . But our sum starts at $n=2$, so its value is $S'=S-(1+x)\le6$ . Can you take it from here ? :-) We have $\dfrac1{1-x}-(1+x)\le6$. Since $|x|<1$, it follows that $1-x>0$, so that multiplying both sides with it won't change the comparison sign. Thus, $1-(1-x^2)\le6(1-x)\to x^2+6x-6$ $\le0$. Now, we know from theory that $ax^2+bx+c$ has the same sign as a outside of the interval $(x_1,x_2)$, and the sign opposite to it inside the same interval, where $x_1$ and $x_2$ are the two roots of the function. In our case, $a=1$, and its roots are $-3\pm\sqrt{15}$ , so $x\in(-3-\sqrt{15},-3+\sqrt{15})$. But, at the same time, $x\in(-1,1)$. Intersecting the two intervals, we have $x\in(-1,-3+\sqrt{15})$, since, obviously, $0<\sqrt{15}<\sqrt{16}=4$.