find the sum of the infinite geometric series 1 + 3/5 + 9/25 + ..., if it exists?

Answer 1

The sum is 5/2.

By the way, the terms in a list like this is called a "sequence." The sum for the sequence is its "series." For any infinite geometric sequence where the common ratio between terms is less than 1, we can find the formula for its series this way:

S = a + ar + ar² + ar^3 + .... + ar^(n-1), where "a" is the first term in the sequence , "r" is the common ratio, and "n" is the number of terms in the sequence.

If we multiply the sum, S, by r, we get this:

Sr = ar + ar² + ar^3 + .... + ar^n. Do you see why the last term here is now ar^n instead of ar^(n - 1)? It's because r [ar^(n-1)] = a [r*r^(n -1)] = a [r^(n -1+1)] = ar^(n - 0) = ar^n.

Now we subtract Sr from S to obtain this:

S - Sr = a + ar + ar² + ar^3 + .... + ar^(n-1) - [ar + ar² + ar^3 + .... + ar^n].

By inspection, we can see that each term drops out except the first term of the first sum and the last term of the second sum. To see this more clearly, you can write this as you might a regular subtraction problem, lining each term in the first sum up with its corresponding term in the second. Then cross each pair out. You will be left with this:

S - Sr = a - ar^n

Now, we solve for S:

S (1 - r) = a (1 - r^n)
S = a [(1 - r^n) / (1 - r)].

Since r < 1, when it is raised to an infinitely high power, the result is essentially zero. The last equation then becomes:

S = a [(1 - 0) / (1 - r)]
S = a [1 / ( 1 - r)].

Now all we have to do is plug your values into this equation to find our answer.

a = 1, r = 3/5 -----> S = 1 [1 / (1 - 3/5)] = 1 [ 1 / (2/5)] = 1 [5/2] = 5/2.