Sequences and Series  The sum of an infinite geometric sequence, infinite geometric series

The sum of an infinite geometric sequence, infinite geometric series
An infinite geometric series converges (has a finite sum even when n is infinitely large) only if the absolute ratio of successive terms is less than 1 that is, if  -1 < r < 1
The sum of an infinite geometric series can be calculated as the value that the finite sum formula takes (approaches) as number of terms n tends to infinity,
 first rewrite Sn, into so that since | r | < 1, then rn ® 0 as  n ® oo
 thus, the sum of an infinite converging geometric series.
The sum of an infinite converging geometric series, examples
Example:  Given a square with side a. Its side is the diagonal of the second square. The side of this square is then the diagonal of the third square and so on, as shows the figure below. Find the sum of areas of all these squares.
Solution:   Using the given conditions,   Example:  Given an equilateral triangle with the side a. Its height is the side of another equilateral triangle. The height of this triangle is then the side of the third equilateral triangle and so on, as shows the figure below. Find the sum of areas of all these triangles.
 Solution:      a1 = h,  a2 = h1,  a3 = h2,  and so on. Thus,   Example:  In a circle of the radius r inscribed is a square, in the square inscribed is another circle and in that circle another square and so on, as is shown in the figure below. Find the sum of areas of all the circles and the sum of areas of all the squares.
Solution:   It follows that the radius r is the half of the diagonal d of the first inscribed square,   Example:  In an isosceles triangle with the base 2a and the angle 2a, opposite to the base, inscribed are  infinite sequence of circles such that first circle touches the base and opposite sides of the triangle while other circles touch opposite sides of the triangle and the preceding circle, as shows the figure below. Find the sum of all circumferences and areas of all inscribed circles.
Solution:   In the shown right triangles we use following relations,   Therefore, the sum of all circumferences, SC = 2pr1 + 2pr2 + 2pr3 + . . . The sum of areas of all circles, SA = r12p + r22p + r32p + . . . Example:  Into the regular square pyramid, with the side of the base a and the height h, inserted is the square with the base lying in the base of the pyramid, and the upper base of which is cross section of the pyramid. Into the space above this square inserted is, the same way, another square, and above of this third square and so on, to infinity, as is shown in the figure below. Find the volume of all inserted squares.
 Solution:   Following relations hold between similar right triangles shown in the figure,   The sum of volumes of all the squares, SV = V1 + V2 + V3  + . . . = a13 + a23 + a33 +  . . .    Intermediate algebra contents 