Prove that 3.999... = 4
(In reply to
By Calculus by K Sengupta)
Consider 0.999..... = 9/10+ 9/(10)^2 +9/(10)^3+......
Since there is an infinite number of terms and the first term of this infinite geometric sequence is 9/10 which is obviously less than 1, we must have:
the sum S= a/(1-r), where a is the first term and r is the common ratio.
Since, (a,r)=(9/10, 1/10), we must have:
S=(9/10)/[1-(1/10)]
= (9/10)/(9/10)
=1
Consequently,:
3.999....
=3+0.999....
=3+1
=4
Puzzle Solution: Method 3
