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My name is Krishna. I'm now in Grade 12. When I was in Grade 11, I saw a question in the math club.The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at theActually, I have already asked a similar question like the one below, but this one is little different.

The question is,

What is the value of, 1 + (1/2) + (1/4) + (1/8) ...... ?

How can you find a definite value for an answer when it keeps on continuing?

Thank you.

Krishna

1etc.,

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it *S*.
Now

so, if we multiply it by 1/2, we getS= 1 + 1/2 + 1/4 + 1/8 + · · ·

(1/2)Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we getS= 1/2 + 1/4 + 1/8 + 1/16 + · · ·

This same technique can be used to find the sum of any "geometric
series", that it, a series where each term is some number *r* times the
previous term. If the first term is *a*, then the series is

so, multiplying both sides byS=a+a r+a r^2 +a r^3 + · · ·

and, subtracting the second equation from the first, you getr S=a r+a r^2 +a r^3 +a r^4 + · · ·

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

then multiply byS=a+a r+a r^2 +a r^3 + · · · +a r^n

and subtract the second from the first, the termsrS=a r+a r^2 +a r^3 +a r^4 + · · · +a r^(n+1)

As long as |*r*| < 1, the term *r*^(*n*+1) will go to zero as *n* goes to
infinity, so the finite sum *S* will approach *a */ (1-*r*) as *n* goes to
infinity. Thus the value of the infinite sum is *a */ (1-*r*), and this also
proves that the infinite sum exists, as long as |*r*| < 1.

In your example, the finite sums were

1 = 2 - 1/1and so on; the

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

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