Geometric Series and Sum to infinity

Question

Given the first two terms of a geometric progression is twice the value of the fifth term, I want to find the least value of n such that $S_n=a(1-r^n)/(1-r)$ is within 5% of $S$. So if we let the first term be $a$ and common ratio be $r$, equating $a + ar = 2/(1-r)$ we can solve for $r= 0.684$ approximately. But how to find $S_n$ within 5% of $S$? Does within means less than or equals to because when I tried to simplify that inequality, I eventually failed to take ln on both sides to solve for n due to getting a negative number..

Answer 1

Here we assume $|r|<1$ as it involes $S_\infty$.

$$a+ar=2\left(\frac{ar^2}{1-r}\right)\\ (1+r)(1-r)=2r^2\\ r=\pm \frac 1{\sqrt{3}}$$ Now we want $$S_n>0.95\ S_\infty\\ \frac {a(1-r^n)}{1-r}>0.95 \ \frac a{1-r}\\ r^n<0.05\\ n>\frac {\log(0.05)}{\log(1/\sqrt{3})}=5.453\qquad (\text{for r>0})\\ n=6\qquad $$

Answer 2

Given ${ar^n}$ its sum is $S=\dfrac{a}{1-r}$

To satisfy the condition that the first two terms are twice the remaining sum we have the equation $$a r+a=2 \left(\frac{a}{1-r}-a-ar\right)$$ which gives (positive) solution $r=\dfrac{1}{\sqrt 3}$

To get the $5\%$ of the sum we solve the equation $$\frac{a(1-r^n)}{1-r}=\frac{5}{100}\frac{a}{1-r}$$ which simplifies to $$r^n=0.05\rightarrow \left(\frac{1}{\sqrt 3}\right)^n<0.05$$ apply logarithm to both sides we get $n\geq 6$

Indeed the sum is $S=\dfrac{3}{3-\sqrt{3}}\approx 2.366$ while $\sum _{n=6}^{\infty } \left(\dfrac{1}{\sqrt{3}}\right)^n\approx 0.08763$ and this value is $3.7\%$ of the whole sum

Hope this helps

PS $n=6$ is valid also if you take the negative ratio $$r=-\dfrac{1}{\sqrt 3}$$ In this case the sum is $S=\dfrac{3}{3+\sqrt{3}}\approx 0.634$ and its $5\%$ is $0.0317$.

Remembering that in alternating series the residue sum when stopping at $n$ is less than $|a_{n+1}|$ we can solve $\left(\dfrac{1}{\sqrt 3}\right)^{n+1}<0.0317$ which gives $n\geq 6$ like before