Infinite Sum of Geometric Series : Part (2)
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.
$|r|>1 \text{ i.e., } r<-1 \text{ or } r>1,$
$(\text{ e.g., } a=2, r=2)$
$\begin{aligned} S_n &= \dfrac{2}{1-2}-\dfrac{2\times2^n}{1-2}\\\\ &= -2+2\times2^n\\\\ &= 2(2^n - 1)\\\\ \end{aligned}$ $\begin{array}{|r|r|} \hline n\hspace{.2cm} & S_n\hspace{4cm} \\ \hline 1 & 2 \\ \hline 5 & 62 \\ \hline 10 & 2046 \\ \hline 50 & 2251799813685246 \\ \hline 100 & 2535301200456458802993406410750 \\ \hline \end{array}$ |
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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။
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။
Case III.
$|r|<1 \text{ i.e., } -1< r < 1 $
$ (\text{ e.g., } a=2, r=\dfrac{1}{2})$
$\begin{aligned} S_n &= \dfrac{2}{1-1/2}-\dfrac{2\times(1/2)^n}{1-1/2}\\\\ &= 4-4{\left( {\dfrac{1}{2}} \right)}^{n}\\\\ &= 4\left[ {1-{{{\left( {\dfrac{1}{2}} \right)}}^{n}}} \right]\\\\ \end{aligned}$ $\begin{array}{|r|l|r|} \hline n\hspace{.2cm} & \hspace{1.5cm}r^n & S_n\hspace{2.5cm} \\ \hline 1 & 0.5 & 2 \\ \hline 2 & 0.25 & 3 \\ \hline 3 & 0.125 & 3.5 \\ \hline 4 & 0.0525 & 3.75 \\ \hline 5 & 0.03125 & 3.875 \\ \hline 10 & 0.0009765625 & 3.9960937500000000000 \\ \hline 50 & 8.88\times 10^{-16}\approx 0 & 3.9999999999999964473 \\ \hline 100 & 7.88\times 10^{-31}\approx 0 & 4.0000000000000000000 \\ \hline 1000 & 9.33\times 10^{-32}\approx 0 & 4.0000000000000000000 \\ \hline \end{array}$ |
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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။
Exercises
- 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 $\dfrac{1}{100}$.
solution
$\begin{aligned}
u_{n} &=\dfrac{1}{3^{n}} \\\\
u_{1} &=\dfrac{1}{3} \\\\
u_{2} &=\dfrac{1}{9} \\\\
\therefore a &=\dfrac{1}{3} \\\\
r=\dfrac{1 / 9}{1 / 3} &=\dfrac{1}{3} \\\\
\end{aligned}$
$\begin{aligned}
S_{n} &=\dfrac{a\left(1-r^{n}\right)}{1-r} \\\\
&=\dfrac{1 / 3\left(1-\left(\dfrac{1}{3}\right)^{n}\right)}{1-\dfrac{1}{3}} \\\\
&=\dfrac{1}{2}\left(1-\left(\dfrac{1}{3}\right)^{n}\right)\\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}} \\\\
&=\dfrac{1}{2} \\\\
\end{aligned}$
$\begin{aligned}
&S -S_{n} <\dfrac{1}{1000} \\\\
&\dfrac{1}{2} -\dfrac{1}{2}\left(1-\left(\dfrac{1}{3}\right)^{n}\right) < \dfrac{1}{1000} \\\\
&\dfrac{1}{2}\left(\dfrac{1}{3}\right)^{n} <\dfrac{1}{1000}\\\\
&\left(\dfrac{1}{3}\right)^{n} <\dfrac{1}{500} \\\\
&3^{n} >500 \\\\
&n >\dfrac{\ln 500}{\ln 3} \\\\
&n >5.66\\\\
\end{aligned}$
The smallest value of $n=6$.
- 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 $\dfrac{1}{1000}$.
solution
$\begin{aligned}
1, \dfrac{1}{5},& \dfrac{1}{25}, \ldots \text { is a G.P. } \\\\
\therefore a &=1 \\\\
r &=\dfrac{1}{5} \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{1}{1-\dfrac{1}{5}} \\\\
&=\dfrac{5}{4}\\\\
S_{n} &=\dfrac{a\left(1-r^{n}\right)}{1-r} \\\\
&=\dfrac{a}{1-r}\left(1-r^{n}\right) \\\\
&=\dfrac{5}{4}\left(1-\left(\dfrac{1}{5}\right)\right)^{n} \\\\
&=\dfrac{5}{4}-\dfrac{5}{4}\left(\dfrac{1}{5}\right)^{n}\\\\
\end{aligned}$
$\begin{aligned}
S-S_{n} & <\dfrac{1}{1000} \\\\
\dfrac{5}{4}\left(\dfrac{1}{5}\right)^{n} & < \dfrac{1}{1000} \\\\
\left(\dfrac{1}{5}\right)^{n} & <\dfrac{1}{1250} \\\\
5^{n} &>1250 \\\\
n &>\dfrac{\ln 1250}{\ln 5} \\\\
n &>4.43\\\\
\end{aligned}$
- 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.
solution
$3(q+5), 3(q+3),(q+7)$ 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.
$\begin{aligned}
&\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{18}{1-\dfrac{2}{3}} \\\\
&=54
\end{aligned}$
- 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 $S$.
solution
$\begin{aligned}
\text{For } 1^{\text{st}} \text{ G.P,}&\\\\
\text{first term } &=7\\\\
\text{common ratio } &=r,\quad |r| < 1\\\\
S&=\dfrac{1}{1-r}\\\\
\text{For } 2^{\text{nd}} \text{ G.P,}&\\\\
\text{first term } &=7\\\\
\text{common ratio } &=\dfrac{r}{4}\\\\
S &=\dfrac{4}{1-\dfrac{r}{4}}\\\\
\end{aligned}$
By the problem,
$\begin{aligned}
&\\
\dfrac{1}{1-r} &=\dfrac{4}{1-\dfrac{r}{4}} \\\\
1-\dfrac{r}{4} &=4-4 r \\\\
4-r &=16-16 r \\\\
15 r &=12 \\\\
r &=\dfrac{4}{5} \\\\
\therefore S &=\dfrac{1}{1-\dfrac{4}{5}} \\\\
&=5
\end{aligned}$
- 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 $r^{n}>0.1$.
solution
$\begin{aligned}
&\text{In a G.P,}\\\\
&\text{ all terms are positive.}\\\\
&\therefore\ a>0\\\\
&\text{common ratio } =r\\\\
& S_{n} < 90 \% \text{ of } S\\\\
& S_{n}<0.9S\\\\
&\dfrac{a\left(1-r^{n}\right)}{1-r} < 0.9 \times \dfrac{a}{1-r}\\\\
&\therefore\ 1-r^{n} < 0.9 \\\\
& r^{n} >1-0.9 \\\\
& r^{n} >0.1
\end{aligned}$
- 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.
solution
$\begin{aligned}
&a=3\left(u_{2}+u_{3}+u_{4}+\cdots\right) \\\\
&a=3(S-a) \\\\
&a=3 S-3 a \\\\
&4 a=3 S \\\\
&4 a=\dfrac{3 a}{1-r} \\\\
&1-r=\dfrac{3}{4} \\\\
&r=\dfrac{1}{4}\\\\
\end{aligned}$
$\begin{aligned}
u_{1}+u_{2} &=15 \\\\
a+a r &=15 \\\\
a(1+r) &=15 \\\\
a\left(1+\dfrac{1}{4}\right) &=15 \\\\
\dfrac{5 a}{4} &=15 \\\\
a &=12 \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{12}{1-\dfrac{1}{4}} \\\\
&=16
\end{aligned}$
- Let $x=1+a+a^{2}+\ldots$ and $y=1+b+b^{2}+\ldots$, where $|a|<1$ and $|b|< 1$. Prove that
$1+a b+a^{2} b^{2}+\ldots=\dfrac{x y}{x+y-1}$.
solution
$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.
- 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.
solution
Let $ u_{1}=a $ and common ratio $=r$
$\begin{aligned}
&\\
\therefore\ a+a r+a r^{2}+\cdots&=15 \\\\
\therefore\ \dfrac{a}{1-r}&=15 \\\\
\therefore a&=15(1-r) \\\\
a+a^{2} r^{2}+a^{2} r^{4}&=45 \\\\
\therefore\ \dfrac{a^{2}}{1-r^{2}}&=45 \\\\
\dfrac{15^{2}(1-r)^{2}}{1-r^{2}}&=45 \\\\
\dfrac{\left(1-r^{2}\right)}{(1-r)(1+r)}&=\dfrac{1}{5}\\\\
\therefore\ \dfrac{1-r}{1+r} &=\dfrac{1}{5} \\\\
1+r &=5-5 r \\\\
6 r &=4 \\\\
r &=\dfrac{2}{3} \\\\
\therefore\ a &=15\left(1-\dfrac{2}{3}\right) \\\\
&=5
\end{aligned}$
- 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 $x$.
solution
$\begin{aligned}
\dfrac{1}{1+x}, & \dfrac{1}{(1+x)^{2}}, \dfrac{1}{(1+x)^{3}}, \ldots \text { is a G.P. } \\\\
\therefore\ a &=\dfrac{1}{1+x} \\\\
&r=\dfrac{\dfrac{1}{(1+x)^{2}}}{\dfrac{1}{1+x}}=\dfrac{1}{1+x}, x \neq-1\\\\
\end{aligned}$
Since the sum to infinity exists,
$\begin{aligned}
&\\
|r| & < 1 \\\\
\left|\dfrac{1}{1+x}\right| & <1 \\\\
|1+x| &>1\\\\
1+ x & <-1 \text { or } 1+x>1 \\\\
x & <-2 \text { or } x>0 \\\\
S&= \dfrac{a}{1-r} \\\\
&= \dfrac{\dfrac{1}{1+x}}{1-\dfrac{1}{1+x}} \\\\
&= \dfrac{1}{1+x} \times \dfrac{1+x}{1+x-1} \\\\
&= \dfrac{1}{x} \text { for }\{x \mid x <-2 \text { or } x>0\}
\end{aligned}$
- If $b=a+a^{2}+a^{3}+\ldots$ where $|a|<1$, prove that $a=\dfrac{b}{1+b}$.
solution
$\begin{aligned}
&b=a+a^{2}+a^{3}+\cdots \text { where }|a| < 1 \\\\
&a, a^{2}, a^{3}, \ldots \text { is a G.P. } \\\\
&\text { st term }=a \\\\
&\text { common ratio }=a \\\\
&\text { Since }|a|<1, \\\\
&\text {Sum to infinity exists.}\\\\
&\therefore\ S=\dfrac{a}{1-r}=\frac{a}{1-a}\\\\
\end{aligned}$
By the problem,
$\begin{aligned}
&\\
S &=b \\\\
\dfrac{a}{1-a} &=b \\\\
a &=b-a b \\\\
a+a b &=b \\\\
a(1+b) &=b \\\\
\therefore\ a &=\dfrac{b}{1+b}
\end{aligned}$
- 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 $y$.
solution
$\begin{aligned}
&\text{Let the given G.P. be}\\\\
&a, a r, a r^{2}, \ldots \text { where }|r| < 1. \\\\
&\therefore\ x=\dfrac{a}{1-r} \\\\
&\therefore\ a=x(1-r)\\\\
&\text{Squaring each terms of given G.P.,}\\\\
&a^{2}, a^{2} r^{2}, a^{2}+\ldots\\\\
\end{aligned}$
$\therefore$ It is also a GP with common ratio $=r^{2}$.
$\begin{aligned}
&\\
\therefore\ y &=\dfrac{a^{2}}{1-r^{2}} \\\\
y &=\dfrac{x^{2}(1-r)^{2}}{(1-r)(1+r)} \\\\
\therefore\ y(1+r) &=x^{2}(1-r) \\\\
y+y r &=x^{2}-x^{2} r \\\\
x^{2} r+y r &=x^{2}-y \\\\
r\left(x^{2}+y\right) &=x^{2}-y \\\\
\therefore\ r &=\dfrac{x^{2}-y}{x^{2}+y}
\end{aligned}$
- 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 $S_{1}+S_{2}+S_{3}+\ldots+S_{p}=\dfrac{p(p+3)}{2}$.
solution
$\begin{aligned}
&S_{1}=\dfrac{1}{1-\dfrac{1}{2}}=2 \\\\
&S_{2}=\dfrac{2}{1-\dfrac{1}{3}}=3 \\\\
&S_{3}=\dfrac{3}{1-\dfrac{1}{4}}=4 \\\\
&\vdots \qquad \qquad \vdots \\\\
&S_{p}=\dfrac{p}{1-\dfrac{1}{p+1}}=p+1\\\\
&\therefore\ S_{1}+S_{2}+S_{3}+\cdots+S_{p}=2+3+4+\cdots+(p+1)\\\\
\end{aligned}$
The terms are in an A.P., with first term $=2$, common difference $=1$.
$\begin{aligned}
&\\
\therefore S_{1}+S_{2}+S_{3}+\cdots+S_{p} &=\dfrac{p}{2}(2(2)+(p-1) 1) \\\\
&=\dfrac{p}{2}(4+p-1) \\\\
&=\dfrac{p(p+3)}{2}\\\\
\end{aligned}$
- A geometric progression has the first term 2 and common ratio $0.95$. Calculate the least value
of $n$ for which $S-S_{n}< 1$.
solution
$\begin{aligned}
&\text{In a G.P.,}\\\\
&a=2 \\\\
&r=0.95 \\\\
&S-S_{n}<1 \\\\
&\dfrac{a}{1-r}-\dfrac{a\left(1-r^{n}\right)}{1-r}<1 \\\\
&\dfrac{a}{1-r}\left(1-1+r^{n}\right)<1\\\\
&\dfrac{2}{1-0.95}(0.95)^{n}<1 \\\\
&\dfrac{2}{0.05}(0.95)^{n}<1 \\\\
&(0.95)^{n}<0.025 \\\\
&n>\dfrac{\ln 0.025}{\ln 0.95} \\\\
&n>71.92\\\\
\end{aligned}$
$\therefore$ The smallest value of $n=72$.
- 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.
solution
$\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} \\\\
- 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.
solution
$\begin{aligned}
\text{In } & \text{ a G.P.}, \\\\
S &=2 a \\\\
\therefore\ \dfrac{a}{1-r}&=2 a \\\\
1-r & =\dfrac{1}{2} \\\\
r & =\dfrac{1}{2} \\\\
u_{5} & =\dfrac{1}{16} \\\\
a r^{4} & =\dfrac{1}{16}\\\\
a\left(\dfrac{1}{2}\right)^{4} &=\dfrac{1}{16} \\\\
\therefore\ a &=1 \\\\
\therefore\ S_{5} &=\dfrac{a\left(1-r^{5}\right)}{1-r} \\\\
&=\dfrac{1\left(1-\left(\dfrac{1}{2}\right)^{5}\right)}{1-\dfrac{1}{2}} \\\\
&=2\left(1-\dfrac{1}{32}\right) \\\\
&=\dfrac{31}{16}
\end{aligned}$
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