The Student Room Group

Reply 1

Just writing it out as a simple series you have
sum from n+1 to inf = sum from 1 to inf - sum from 1 to n
or equivalently as per the previous comment
sum from 1 to n + sum from n+1 to inf = sum from 1 to inf

Its one way to derive the sum to n formula of a geometric series, based on the infinite sum. The numerator is a - ar^(n+1) which represents
sum from 1 to inf - sum from n+1 to inf
as ar^(n+1) is the first term in the geometric series that starts at n+1.
(edited 2 years ago)

Reply 2

Original post by mqb2766
Just writing it out as a simple series you have
sum from n+1 to inf = sum from 1 to inf - sum from 1 to n
or equivalently as per the previous comment
sum from 1 to n + sum from n+1 to inf = sum from 1 to inf

Its one way to derive the sum to n formula of a geometric series, based on the infinite sum. The numerator is a - ar^(n+1) which represents
sum from 1 to inf - sum from n+1 to inf
as ar^(n+1) is the first term in the geometric series that starts at n+1.

So sum from n+1 to inf = sum from 1 to inf - sum from 1 to n is always true for any summations?

Reply 3

Original post by action123
So sum from n+1 to inf = sum from 1 to inf - sum from 1 to n is always true for any summations?

You could have written down a simple example but f the series was the sum of natural numbers (ignoring the fact that the infinite sum doesnt converge...) then the sum from 1 to infinity is
1 + 2 + 3 +....+ n + (n+1) + (n+2) + ...
Obviously you can write this as
[1 + 2 + 3+....+n] + [(n+1) + (n+2) + ... ]
so
sum from 1 to n + sum from (n+1) to inf = sum from 1 to inf

You shouldnt have to remember it, it drops out of the definition of the series (splitting at n/n+1). Assuming the sum to infinity converges, then its true for any series.
(edited 2 years ago)