Infinite Sum of Geometric Series : Part (2)

Infinite Geometric Series

A geometric series with infinite number of terms is called an infinite geometric series.

Geometric Progression တစ်ခု၏ $n$ terms အထိ ပေါင်းလဒ်ကို

$S_n= \dfrac{a(1-r^n)}{1-r}$

ဟုသိရှိခဲ့ပြီး ဖြစ်သည်။ ညီမျှခြင်းကို အောက်ပါအတိုင်း အကျယ်ဖြန့်ကြည့်ပါမည်။

$\begin{aligned} S_n&= \dfrac{a(1-r^n)}{1-r}\\\\ &= \dfrac{a-ar^n}{1-r}\\\\ &= \dfrac{a}{1-r}-\dfrac{ar^n}{1-r}\\\\ \end{aligned}$

Case I.
$|r|>1 \text{ i.e., } r<-1 \text{ or } r>1,$
$(\text{ e.g., } a=2, r=2)$

$\begin{aligned} S_n &= \dfrac{2}{1-2}-\dfrac{2\times2^n}{1-2}\\\\ &= -2+2\times2^n\\\\ &= 2(2^n - 1)\\\\ \end{aligned}$ $\begin{array}{|r|r|} \hline n\hspace{.2cm} & S_n\hspace{4cm} \\ \hline 1 & 2 \\ \hline 5 & 62 \\ \hline 10 & 2046 \\ \hline 50 & 2251799813685246 \\ \hline 100 & 2535301200456458802993406410750 \\ \hline \end{array}$

ထိုအခါ $n$ ၏ တန်ဖိုးကြီးလာသည်နှင့် အမျှ $S_n$ ၏ ပမာဏ (magnitude) သည်လည်း ကြီးလာပါမည်။

$n\rightarrow \infty$ ($n$ သည် အနန္တပမာဏသို့ ချဉ်းကပ်သွားသည်)

$S_n\rightarrow \infty$ ($S_n$ သည်လည်း အနန္တပမာဏသို့ ချဉ်းကပ်သွားသည်)

ထို့ကြောင့် အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်ကို ရှာယူရန် မဖြစ်နိုင်တော့ပါ။

ထိုအခြေအနေကို divergent condition ဟုခေါ်သည်။ Divergent Condition တွင် Infinite Sum (အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်) ကို ရှာ၍မရနိုင်ပါ။

Key Point

If $|r|>1$, the geometric series is divergent, and Sum to Infinity doesn't exist.

Case II.
$|r|=1 \text{ i.e., } r=\pm 1$

အကယ်၍ $r=1$ ဖြစ်လျှင်

$S_n = a + a + a + \dots$ to $n$ terms = $na$

ထိုအခါ $n$ ၏ တန်ဖိုးကြီးလာသည်နှင့် အမျှ $S_n$ ၏ ပမာဏ (magnitude) သည်လည်း ကြီးလာပါဦးမည်။

ထို့ကြောင့် အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်ကို ရှာယူရန် မဖြစ်နိုင်တော့ပါ။

အကယ်၍ $r=-1$ ဖြစ်လျှင်

$S_n = a - a + a -a + a - \dots$ to $n$ terms = $a \text { if } n \text { is odd}$

$S_n = a - a + a -a + a - \dots$ to $n$ terms = $0 \text { if } n \text { is even}$

ထို့ကြောင့် အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်ကို ရှာယူရန် မလိုတော့ပါ။

Key Point

If $|r|=1$,the geometric series is also a type of divergent series and Sum to Infinity doesn't exist.

Case III.
$|r|<1 \text{ i.e., } -1< r < 1 $
$ (\text{ e.g., } a=2, r=\dfrac{1}{2})$

$\begin{aligned} S_n &= \dfrac{2}{1-1/2}-\dfrac{2\times(1/2)^n}{1-1/2}\\\\ &= 4-4{\left( {\dfrac{1}{2}} \right)}^{n}\\\\ &= 4\left[ {1-{{{\left( {\dfrac{1}{2}} \right)}}^{n}}} \right]\\\\ \end{aligned}$ $\begin{array}{|r|l|r|} \hline n\hspace{.2cm} & \hspace{1.5cm}r^n & S_n\hspace{2.5cm} \\ \hline 1 & 0.5 & 2 \\ \hline 2 & 0.25 & 3 \\ \hline 3 & 0.125 & 3.5 \\ \hline 4 & 0.0525 & 3.75 \\ \hline 5 & 0.03125 & 3.875 \\ \hline 10 & 0.0009765625 & 3.9960937500000000000 \\ \hline 50 & 8.88\times 10^{-16}\approx 0 & 3.9999999999999964473 \\ \hline 100 & 7.88\times 10^{-31}\approx 0 & 4.0000000000000000000 \\ \hline 1000 & 9.33\times 10^{-32}\approx 0 & 4.0000000000000000000 \\ \hline \end{array}$

ထိုအခါ $n$ ၏ တန်ဖိုးကြီးလာသည်နှင့် အမျှ $S_n$ ၏ ပမာဏ (magnitude) သည် တန်ဖိုးတစ်ခုတွင် မပြောင်းလဲတော့ဘဲ ကိန်းသေ ဖြစ်လာပါသည်။

$|r|< 1$ အတွက် $n\rightarrow \infty$ ဖြစ်သည့်အခါ $S_n$ သည် တန်ဖိုးတစ်ခုတွင် မပြောင်းမလဲ တည်ရှိနေသည်။ အဆိုပါတန်ဖိုးမှာ limit of $S_n$ as $n$ approaches $\infty$ ပင် ဖြစ်သည်။

$\begin{aligned} \lim _{n \rightarrow \infty} S_{n} &=\lim _{n \rightarrow \infty}\left(\dfrac{a}{1-r}-\dfrac{a r^{n}}{1-r}\right) \\ &=\dfrac{a}{1-r} \times \lim _{n \rightarrow \infty}\left(1-r^{n}\right) \\ &=\dfrac{a}{1-r}(1-0) \quad(\because\ |r|<1)\\ &=\dfrac{a}{1-r} \end{aligned}$

အထက်ဖေါ်ပြပါ geometric series အမျိုးအစားကို convergent series ဟုခေါ်ပြီး Converget Geometric Series တစ်ခုတွင် infinite sum (Sum to infinity) ရှိသည်ဟု မှတ်ယူရမည်။ Sum to Infinity ကို သင်္ကေတအားဖြင့် $(S)$ ဟုသတ်မှတ်သည်။

Key Point

If $|r|<1$,the geometric series is convergent series and Sum to Infinity exists. The sum to infinity of a convergent geometric series is denoted by $S$.

$\begin{array}{|l|} \hline S=\dfrac{a}{1-r}\\ \hline \end{array}$

Exercises

1. A geometric progression is defined by $u_{n}=\dfrac{1}{3^{n}}$. Find $S_{n}$ and the smallest value of $n$ for which the sum of the first $n$ terms and the sum to infinity differ by less than $\dfrac{1}{100}$.
   
   
   
   

   $\begin{aligned} u_{n} &=\dfrac{1}{3^{n}} \\\\ u_{1} &=\dfrac{1}{3} \\\\ u_{2} &=\dfrac{1}{9} \\\\ \therefore a &=\dfrac{1}{3} \\\\ r=\dfrac{1 / 9}{1 / 3} &=\dfrac{1}{3} \\\\ \end{aligned}$ $\begin{aligned} S_{n} &=\dfrac{a\left(1-r^{n}\right)}{1-r} \\\\ &=\dfrac{1 / 3\left(1-\left(\dfrac{1}{3}\right)^{n}\right)}{1-\dfrac{1}{3}} \\\\ &=\dfrac{1}{2}\left(1-\left(\dfrac{1}{3}\right)^{n}\right)\\\\ S &=\dfrac{a}{1-r} \\\\ &=\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}} \\\\ &=\dfrac{1}{2} \\\\ \end{aligned}$ $\begin{aligned} &S -S_{n} <\dfrac{1}{1000} \\\\ &\dfrac{1}{2} -\dfrac{1}{2}\left(1-\left(\dfrac{1}{3}\right)^{n}\right) < \dfrac{1}{1000} \\\\ &\dfrac{1}{2}\left(\dfrac{1}{3}\right)^{n} <\dfrac{1}{1000}\\\\ &\left(\dfrac{1}{3}\right)^{n} <\dfrac{1}{500} \\\\ &3^{n} >500 \\\\ &n >\dfrac{\ln 500}{\ln 3} \\\\ &n >5.66\\\\ \end{aligned}$ The smallest value of $n=6$.

   
   
   
2. Find the smallest value of $n$ for which the sum to $\mathrm{n}$ terms and the sum to infinity of a G.P. $1, \dfrac{1}{5}, \dfrac{1}{25}, \ldots$ differ by less than $\dfrac{1}{1000}$.
   
   
   
   

   $\begin{aligned} 1, \dfrac{1}{5},& \dfrac{1}{25}, \ldots \text { is a G.P. } \\\\ \therefore a &=1 \\\\ r &=\dfrac{1}{5} \\\\ S &=\dfrac{a}{1-r} \\\\ &=\dfrac{1}{1-\dfrac{1}{5}} \\\\ &=\dfrac{5}{4}\\\\ S_{n} &=\dfrac{a\left(1-r^{n}\right)}{1-r} \\\\ &=\dfrac{a}{1-r}\left(1-r^{n}\right) \\\\ &=\dfrac{5}{4}\left(1-\left(\dfrac{1}{5}\right)\right)^{n} \\\\ &=\dfrac{5}{4}-\dfrac{5}{4}\left(\dfrac{1}{5}\right)^{n}\\\\ \end{aligned}$ $\begin{aligned} S-S_{n} & <\dfrac{1}{1000} \\\\ \dfrac{5}{4}\left(\dfrac{1}{5}\right)^{n} & < \dfrac{1}{1000} \\\\ \left(\dfrac{1}{5}\right)^{n} & <\dfrac{1}{1250} \\\\ 5^{n} &>1250 \\\\ n &>\dfrac{\ln 1250}{\ln 5} \\\\ n &>4.43\\\\ \end{aligned}$

   
   
   
3. The first three terms of a geometric progression are $3(q+5), 3(q+3)$ and $(q+7)$ respectively. Calculate the possible values of $q$. For each possible value of $q$ find the common ratio and the sum to infinity of the geometric progression.
   
   
   
   

   $3(q+5), 3(q+3),(q+7)$ are the first three terms of a G.P. $\begin{aligned} &\\ &\therefore \dfrac{q+3}{q+5}=\dfrac{q+7}{3(q+3)} \\\\ &3\left(q^{2}+6 q+9\right)=q^{2}+12 q+35 \\\\ &2 q^{2}+6 q-8=0 \\\\ &q^{2}+3 q-4=0 \\\\ &(q+4)(q-1)=0\\\\ &\therefore q=-4 \text{ or } q=1\\\\ &\text{When } q=-4\\\\ &a=u_{1}=3(-4+5)=3 \\\\ &r=\dfrac{q+3}{q+5}=\dfrac{-4+3}{-4+5}=-1 \\\\ &|r|=1\\\\ \end{aligned}$ $\therefore$ Sum to infinity does not exist. $\begin{aligned} &\\ &\text{When } q=1\\\\ &a=u_{1}=3(1+5)=18 \\\\ &r=\dfrac{q+3}{q+5}=\dfrac{1+3}{1+5}=\dfrac{2}{3}<1\\\\ \end{aligned}$ $\therefore$ Sum to infinity exists. $\begin{aligned} &\\ S &=\dfrac{a}{1-r} \\\\ &=\dfrac{18}{1-\dfrac{2}{3}} \\\\ &=54 \end{aligned}$

   
   
   
4. A geometric progression has first term $1$ and common ratio $r$. A second geometric progression has first term $4$ and common ratio $\dfrac{r}{4}$. The two progressions have the same sum to infinity, $S$. Find the values of $r$ and $S$.
   
   
   
   

   $\begin{aligned} \text{For } 1^{\text{st}} \text{ G.P,}&\\\\ \text{first term } &=7\\\\ \text{common ratio } &=r,\quad |r| < 1\\\\ S&=\dfrac{1}{1-r}\\\\ \text{For } 2^{\text{nd}} \text{ G.P,}&\\\\ \text{first term } &=7\\\\ \text{common ratio } &=\dfrac{r}{4}\\\\ S &=\dfrac{4}{1-\dfrac{r}{4}}\\\\ \end{aligned}$ By the problem, $\begin{aligned} &\\ \dfrac{1}{1-r} &=\dfrac{4}{1-\dfrac{r}{4}} \\\\ 1-\dfrac{r}{4} &=4-4 r \\\\ 4-r &=16-16 r \\\\ 15 r &=12 \\\\ r &=\dfrac{4}{5} \\\\ \therefore S &=\dfrac{1}{1-\dfrac{4}{5}} \\\\ &=5 \end{aligned}$

   
   
   
5. A geometric progression, in which all the terms are positive, has common ratio $r$. The sum of the first $n$ terms is less than $90 \%$ of the sum to infinity. Show that $r^{n}>0.1$.
   
   
   
   

   $\begin{aligned} &\text{In a G.P,}\\\\ &\text{ all terms are positive.}\\\\ &\therefore\ a>0\\\\ &\text{common ratio } =r\\\\ & S_{n} < 90 \% \text{ of } S\\\\ & S_{n}<0.9S\\\\ &\dfrac{a\left(1-r^{n}\right)}{1-r} < 0.9 \times \dfrac{a}{1-r}\\\\ &\therefore\ 1-r^{n} < 0.9 \\\\ & r^{n} >1-0.9 \\\\ & r^{n} >0.1 \end{aligned}$

   
   
   
6. In an infinite G.P, each term is equal to three times the sum of all the terms that follow it. The sum of the first two terms is $15$ . Find the sum of the series to infinity.
   
   
   
   

   $\begin{aligned} &a=3\left(u_{2}+u_{3}+u_{4}+\cdots\right) \\\\ &a=3(S-a) \\\\ &a=3 S-3 a \\\\ &4 a=3 S \\\\ &4 a=\dfrac{3 a}{1-r} \\\\ &1-r=\dfrac{3}{4} \\\\ &r=\dfrac{1}{4}\\\\ \end{aligned}$ $\begin{aligned} u_{1}+u_{2} &=15 \\\\ a+a r &=15 \\\\ a(1+r) &=15 \\\\ a\left(1+\dfrac{1}{4}\right) &=15 \\\\ \dfrac{5 a}{4} &=15 \\\\ a &=12 \\\\ S &=\dfrac{a}{1-r} \\\\ &=\dfrac{12}{1-\dfrac{1}{4}} \\\\ &=16 \end{aligned}$

   
   
   
7. Let $x=1+a+a^{2}+\ldots$ and $y=1+b+b^{2}+\ldots$, where $|a|<1$ and $|b|< 1$. Prove that $1+a b+a^{2} b^{2}+\ldots=\dfrac{x y}{x+y-1}$.
   
   
   
   

   $x=1+a+a^{2}+\cdots,\quad |a|< 1\\\\ $ Given terms are in G.P. with ist term $=1$ and the common ratio $=a$ $\begin{aligned} &\\ &\therefore x=\dfrac{1}{1-a} \\\\ &y=1+b+b^{2}+\cdots,\quad |b|<1\\\\ \end{aligned}$ Given terms are in G.P. with ist term $=1$ and the common ratio $=b$ $\begin{aligned} &\\ \therefore y=\dfrac{1}{1-b}\\\\ \end{aligned}$ Let $S=1+a b+a^{2} b^{2}+\cdots\\\\ $ The terms are also in G.P., with first term $=1$ and common ratio $=a b$. $\begin{aligned} &\\ \therefore\ S&=\dfrac{1}{1-a b}\\\\ x y &=\dfrac{1}{1-a} \cdot \dfrac{1}{1-b} \\\\ &=\dfrac{1}{1-a-b+a b} \\\\ x+y-1 &=\dfrac{1}{1-a}+\dfrac{1}{1-b}-1 \\\\ &=\dfrac{1-b+1-a-1+a+b-a b}{1-a-b+a b} \\\\ &=\dfrac{1-a b}{1-a-b+a b}\\\\ \dfrac{x y}{x+y-1} &=\dfrac{1}{1-a b} \\\\ &=S\\\\ \end{aligned}$ Hence, proved.

   
   
   
8. The sum of an infinite geometric series is $15$ and the sum of the squares of the terms to infinity is $45$ . Find the first term and the common ratio.
   
   
   
   

   Let $ u_{1}=a $ and common ratio $=r$ $\begin{aligned} &\\ \therefore\ a+a r+a r^{2}+\cdots&=15 \\\\ \therefore\ \dfrac{a}{1-r}&=15 \\\\ \therefore a&=15(1-r) \\\\ a+a^{2} r^{2}+a^{2} r^{4}&=45 \\\\ \therefore\ \dfrac{a^{2}}{1-r^{2}}&=45 \\\\ \dfrac{15^{2}(1-r)^{2}}{1-r^{2}}&=45 \\\\ \dfrac{\left(1-r^{2}\right)}{(1-r)(1+r)}&=\dfrac{1}{5}\\\\ \therefore\ \dfrac{1-r}{1+r} &=\dfrac{1}{5} \\\\ 1+r &=5-5 r \\\\ 6 r &=4 \\\\ r &=\dfrac{2}{3} \\\\ \therefore\ a &=15\left(1-\dfrac{2}{3}\right) \\\\ &=5 \end{aligned}$

   
   
   
9. Express the values of $x$ for which the sum to infinity of an infinite geometric series $\dfrac{1}{1+x}+\dfrac{1}{(1+x)^{2}}+\dfrac{1}{(1+x)^{3}}+\ldots$ exists. Hence find the general expression for the sum to infinity for this range of $x$.
   
   
   
   

   $\begin{aligned} \dfrac{1}{1+x}, & \dfrac{1}{(1+x)^{2}}, \dfrac{1}{(1+x)^{3}}, \ldots \text { is a G.P. } \\\\ \therefore\ a &=\dfrac{1}{1+x} \\\\ &r=\dfrac{\dfrac{1}{(1+x)^{2}}}{\dfrac{1}{1+x}}=\dfrac{1}{1+x}, x \neq-1\\\\ \end{aligned}$ Since the sum to infinity exists, $\begin{aligned} &\\ |r| & < 1 \\\\ \left|\dfrac{1}{1+x}\right| & <1 \\\\ |1+x| &>1\\\\ 1+ x & <-1 \text { or } 1+x>1 \\\\ x & <-2 \text { or } x>0 \\\\ S&= \dfrac{a}{1-r} \\\\ &= \dfrac{\dfrac{1}{1+x}}{1-\dfrac{1}{1+x}} \\\\ &= \dfrac{1}{1+x} \times \dfrac{1+x}{1+x-1} \\\\ &= \dfrac{1}{x} \text { for }\{x \mid x <-2 \text { or } x>0\} \end{aligned}$

   
   
   
10. If $b=a+a^{2}+a^{3}+\ldots$ where $|a|<1$, prove that $a=\dfrac{b}{1+b}$.
    
    
    
    

    $\begin{aligned} &b=a+a^{2}+a^{3}+\cdots \text { where }|a| < 1 \\\\ &a, a^{2}, a^{3}, \ldots \text { is a G.P. } \\\\ &\text { st term }=a \\\\ &\text { common ratio }=a \\\\ &\text { Since }|a|<1, \\\\ &\text {Sum to infinity exists.}\\\\ &\therefore\ S=\dfrac{a}{1-r}=\frac{a}{1-a}\\\\ \end{aligned}$ By the problem, $\begin{aligned} &\\ S &=b \\\\ \dfrac{a}{1-a} &=b \\\\ a &=b-a b \\\\ a+a b &=b \\\\ a(1+b) &=b \\\\ \therefore\ a &=\dfrac{b}{1+b} \end{aligned}$

    
    
    
11. The sum of infinite terms of a G.P. is $x$ and on squaring each term of it, the sum will be $y$, find the common ratio in terms of $x$ and $y$.
    
    
    
    

    $\begin{aligned} &\text{Let the given G.P. be}\\\\ &a, a r, a r^{2}, \ldots \text { where }|r| < 1. \\\\ &\therefore\ x=\dfrac{a}{1-r} \\\\ &\therefore\ a=x(1-r)\\\\ &\text{Squaring each terms of given G.P.,}\\\\ &a^{2}, a^{2} r^{2}, a^{2}+\ldots\\\\ \end{aligned}$ $\therefore$ It is also a GP with common ratio $=r^{2}$. $\begin{aligned} &\\ \therefore\ y &=\dfrac{a^{2}}{1-r^{2}} \\\\ y &=\dfrac{x^{2}(1-r)^{2}}{(1-r)(1+r)} \\\\ \therefore\ y(1+r) &=x^{2}(1-r) \\\\ y+y r &=x^{2}-x^{2} r \\\\ x^{2} r+y r &=x^{2}-y \\\\ r\left(x^{2}+y\right) &=x^{2}-y \\\\ \therefore\ r &=\dfrac{x^{2}-y}{x^{2}+y} \end{aligned}$

    
    
    
12. If $S_{1}, S_{2}, S_{3}, \ldots, S_{p}$ are the sums to infinity of geometric series whose first terms are $1,2,3, \ldots, p$ and whose common ratios are $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \ldots, \dfrac{1}{p+1}$ respectively, show that $S_{1}+S_{2}+S_{3}+\ldots+S_{p}=\dfrac{p(p+3)}{2}$.
    
    
    
    

    $\begin{aligned} &S_{1}=\dfrac{1}{1-\dfrac{1}{2}}=2 \\\\ &S_{2}=\dfrac{2}{1-\dfrac{1}{3}}=3 \\\\ &S_{3}=\dfrac{3}{1-\dfrac{1}{4}}=4 \\\\ &\vdots \qquad \qquad \vdots \\\\ &S_{p}=\dfrac{p}{1-\dfrac{1}{p+1}}=p+1\\\\ &\therefore\ S_{1}+S_{2}+S_{3}+\cdots+S_{p}=2+3+4+\cdots+(p+1)\\\\ \end{aligned}$ The terms are in an A.P., with first term $=2$, common difference $=1$. $\begin{aligned} &\\ \therefore S_{1}+S_{2}+S_{3}+\cdots+S_{p} &=\dfrac{p}{2}(2(2)+(p-1) 1) \\\\ &=\dfrac{p}{2}(4+p-1) \\\\ &=\dfrac{p(p+3)}{2}\\\\ \end{aligned}$

    
    
    
13. A geometric progression has the first term 2 and common ratio $0.95$. Calculate the least value of $n$ for which $S-S_{n}< 1$.
    
    
    
    

    $\begin{aligned} &\text{In a G.P.,}\\\\ &a=2 \\\\ &r=0.95 \\\\ &S-S_{n}<1 \\\\ &\dfrac{a}{1-r}-\dfrac{a\left(1-r^{n}\right)}{1-r}<1 \\\\ &\dfrac{a}{1-r}\left(1-1+r^{n}\right)<1\\\\ &\dfrac{2}{1-0.95}(0.95)^{n}<1 \\\\ &\dfrac{2}{0.05}(0.95)^{n}<1 \\\\ &(0.95)^{n}<0.025 \\\\ &n>\dfrac{\ln 0.025}{\ln 0.95} \\\\ &n>71.92\\\\ \end{aligned}$ $\therefore$ The smallest value of $n=72$.

    
    
    
14. In an infinite G.P., if the first term is equal to the twice of the sum of the remaining terms, find the common ratio.
    
    
    
    

    $\begin{aligned} \text{In } & \text{ a G.P.} \\\\ a&= 2\left(u_{2}+u_{3}+u_{4}+\cdots\right) \\\\ a&= 2(S-a) \\\\ a&= 2 S-2 a \\\\ 3 a&= 2 S \\\\ 3 a&= \dfrac{2 a}{1-r} \\\\ 1-r&= \dfrac{2}{3} \\\\ r&= \dfrac{1}{3} \end{aligned}$

    
    
    
15. If the sum to infinity of an infinite geometric progression is twice the first term and the fifth term is $\dfrac{1}{16}$, find the sum of the first five terms of that G.P.
    
    
    
    

    $\begin{aligned} \text{In } & \text{ a G.P.}, \\\\ S &=2 a \\\\ \therefore\ \dfrac{a}{1-r}&=2 a \\\\ 1-r & =\dfrac{1}{2} \\\\ r & =\dfrac{1}{2} \\\\ u_{5} & =\dfrac{1}{16} \\\\ a r^{4} & =\dfrac{1}{16}\\\\ a\left(\dfrac{1}{2}\right)^{4} &=\dfrac{1}{16} \\\\ \therefore\ a &=1 \\\\ \therefore\ S_{5} &=\dfrac{a\left(1-r^{5}\right)}{1-r} \\\\ &=\dfrac{1\left(1-\left(\dfrac{1}{2}\right)^{5}\right)}{1-\dfrac{1}{2}} \\\\ &=2\left(1-\dfrac{1}{32}\right) \\\\ &=\dfrac{31}{16} \end{aligned}$

    
    
    

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