# Finding minimum of geometric sequence

For the given natural number $$0, define the function as follows:

$$f\left(r\right)=2\cdot\sum_{i=0}^{n+2}r^{i-n}+1$$

I want to find the minimum for that function in the domain $$r\in\left(0,\infty\right)$$, I type some parameters in demos and it seems that the minimum is equal to $$r=1.5$$

But I don't know how to prove it formally, and why it's not a function of $$n$$

• It is a function of $n$. For example, what if $n=0$? Then the minimum will be at $r=-.5$. Dec 2, 2022 at 18:47
• I stated that n is natural number so it should be greater then 0 Dec 2, 2022 at 18:51
• The naturals include zero, as a definition is common.
– Nij
Dec 2, 2022 at 19:47

Evaluating the geometric sum: $$f(r)=2\cdot\frac{r^{-n}-r^3}{1-r}+1$$ Note that $$f(1)=n+3$$. This function is extremized when the function $$g(r)=\frac{r^{-n}-r^3}{1-r}$$ is extremized. Let's work with that instead. Differentiating and equating to zero: $$\frac{\mathrm{d}g}{\mathrm{d}r}=\frac{r^{-1-n}\left(n(r-1)+r+r^{3+n}(2r-3)\right)}{(1-r)^2}=0$$ Simplify, using $$r>0$$ and $$n>0$$: $$\implies n(r-1)+r+r^{3+n}(2r-3)=0$$ The minimum may be at the root of this equation. For example, for $$n=3$$, the minimum is at $$r=1.1908$$. Examining this interactive Desmos graph confirms this is true. Playing around with it also shows that the minimum is not $$r=1.5$$, and it really does depend on $$n$$.
I will note that, in the limit $$n\to\infty$$, $$g(r)\to-\frac{r^3}{1-r}$$ so our optimization condition becomes: $$\frac{r^2(2r-3)}{(r-1)^2}=0$$ For $$r>0$$, this becomes $$2r-3=0\Longleftrightarrow r=\frac{3}{2}$$. So, in the limit $$n\to\infty$$, the sum is minimized at $$r=\frac{3}{2}$$.