Geometric Series Problem Help! Thank you.

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+1/10+...
Find the values of S1, S2, S3


What is the value of S10?
And S2016?
The values 1/s1,1/s2+1/s3+1/s4+1/s5?
And does 1/s1+1/s2+....(infinite) have a sum? Explain.

Answer 1

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 fisrt 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→∝, [ar^n/(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

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Sn=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/Sn=n/n+1 and the sum consists of consecutive  terms viz., 1/S(n-1)=(n-1)/n, 1/Sn=n/n+1 and 1/S(n+1)=n+1/n+2 and 1/S(n+1)>1/Sn>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.