Question

show that in an infinite G.p each terms bears a constant ratio to the sum of all the terms that follow it

Answer 1

An infinite G.P is a sequence of the form:

$a,ar,ar^{2},ar^{3} ,...ar^{n-1},...$

ith term in G.P = $ar^{i-1}$

sum of all terms after ith term,

$S_i=\lim_{x \to\infty}\sum\limits^x_{k=i+1} {ar^{k-1}}\\=a\lim_{x \to\infty}\sum\limits^x_{k=i+1} {r^{k-1}}\\\\=a\lim_{x \to\infty}(\sum\limits^x_{k=1} {r^{k-1}}-\sum\limits^i_{k=1} {r^{k-1}})\\=a\lim_{x \to\infty}(\frac{r^x-1}{r-1} -\frac{r^i-1}{r-1})\\=a\lim_{x \to\infty}\frac{r^\infty-1-r^i+1}{r-1}\\=a\lim_{x \to\infty}\frac{r^\infty-r^i}{r-1}$

Now, ratio of ith term to that of the sum of all the terms that follow it

$=\frac{ar^{i-1}}{a\lim_{x \to\infty}\frac{r^\infty-r^i}{r-1}}\\\\=\frac{r^{i-1}}{\lim_{x \to\infty}\frac{r^x-r^i}{r-1}}\\\\=\frac{r^{i-1}(r-1)}{\lim_{x \to\infty}(r^x-r^i)}\\=\frac{r^{i-1}(r-1)}{r^{i-1}\lim_{x \to\infy}(r^{x-i+1}-r)}\\=\frac{r-1}{\lim_{x \to\infty}(r^{x-i+1}-r)}\\=\frac{r-1}{\lim_{x \to\infty}(r^{x-i+1})-r}\\ \implies\frac{r-1}{\lim_{x \to\infty}(r^{x-i+1})-r}=\frac{r-1}{r^{\infty}-r}=0$

We see that it does not bear a constant ratio, instead the ratio tends to 0

because infinity (∞) is not a constant.