Infinite Geometric Series

Question

Consider the following Infinite Geometric Series:

S₁=1+1/2+1/4+1/8...

 

S₂=1+1/3+1/9+...

 

S₃=1+1/4+1/16+...

 

-Find the values of S_1, S_2, S₃ 

-What is the value of S_10? And S₂₀₁₆?

-The values of 1/S₁, 1/S₂, 1/S₃ form another sequence of its own.

What is the value of 1/S_{1 +} 1/S₂ + 1/S₃ + 1/S₄ + 1/S₅? 

And does 1/S₁ + 1/S₂+...(infinite) have a sum? Explain.

 

 

Answer 1

This is an answer to the last part of the question.

 

As Sanhita has described, S_n = 1 + 1/n, n = 1, 2, 3, ...

 

Now n < n +1 => 1/(n + 1) < 1/n

                      => 1 + 1/(n+1) < 1 + 1/n

                      => S_n+1 < S_n

                       => 1/S_n < 1/S_n+1

 

Now 1/S_n = n/(n+1) and

 

lim (n→∞) 1/S_n = 1

 

∴ The sum of 1/S₁ + 1/S₂ + 1/S₃ .... → ∞

because each new term added to the series is larger than the

previous term and each also gets closer to 1.  This means that

each new term added to the series increases the sum of the series.

 

There is no finite sum for this series.

 

Answer 2

S₁, S₂, S₃ are sum of infinite G.P series with common ratio having values between -1.0 to +1.0

Since any G.P series, with first term =a and common ratio=r may have sum, S= a+ar+ar²+ar³+....+ar^n-1 ......... (1) if the series has finite numbers of terms =n where n is an integer.
Multiplying both side in (1) by r,
Sr=ar+ar²+ar³+....+ar^n-1+arⁿ ..........(2)
Subtracting (2) from (1)
S-Sr=a-arⁿ
=>S(1-r)=a(1-rⁿ)
=<S=a(1-rⁿ⁾/(1-r)=[a/(1-r)]-[arⁿ/(1-r)],
where r≠0 and -1<r<1 and n→∝, [arⁿ/(1-r)]→0
Thus, sum of infinite G.P. series with common ratio having values between -1.0 to +1.0, S=a/(1-r)
Given that,
S₁=1+1/2+1/4+1/8... =1/(1-1/2)=1/{1/2}=2
S₂=1+1/3+1/9+...=1/(1-1/3)=1/(2/3}=3/2
S₃=1+1/4+1/16+...=1/(1-1/4)=1/{3/4}=4/3
.
.
.
S_n=1+1/(n+1)+1/(n+1)²+...... =1/[1-1/(n+1)]=1/{(n+1-1)/n+1}=(n+1)/n=1+1/n, where n is an integer
Thus, S₁₀=1+1/11+1/121+...=1+1/10=11/10
S₂₀₁₆ =1+1/2017+1/(2017)²+....=1+1/2016=2017/2016

S₄=1+1/5+1/25+...=1/(1-1/5)=5/4
S₅=1+1/6+1/36+...=1/(1-1/6)=6/5

1/S₁+1/S₂+ 1/S₃ +1/S₄ + 1/ S₅=1/2+2/3+3/4+4/5+5/6=[30+20+15+12+10]/60=87/60=29/20

Yes, off course 1/S₁+1/S₂+....(infinite) has a sum as we see, 1/S_n=n/n+1 and the sum consists of consecutive terms viz., 1/S_(n-1)=(n-1)/n, 1/S_n=n/n+1 and 1/S_(n+1)=n+1/n+2 and 1/S_(n+1)>1/S_n>1/S_(n-1) where n is an integer
If n→∝, then (n+1)→∝ and also [n/n+1 ]=undefined . Therefore sum of 1/S1+1/S2+....(infinite) remains undefined.