# Math Help - Archimedean series

1. ## Archimedean series

This problem has left me scratching my head:
The modern (and Archimedean!) meaning of “the series Σ ai (where i goes from 0 to ∞)converges to A" isusually captured by a definition like:

(*)Σ ai (where i goes from 0 to ∞)converges to A if for every ε > 0 there is a K such that for all k K wehave (Σ ai) – A │< ε (where i goes from 0 to k).

Archimedes himself would probably have said something more along the following lines:

(º) Σ ai (where i goes from 0 to ∞)converges to A if both

(1) for every L < A there is a K such that for all k K we have L < (Σ ai) (where i goes from 0 to k),
and

(2) for every U > A there is a Ksuch that for all k Kwe have (Σ ai) < U(where i goes from 0 to k).

Explain, in detail, why these two definitions are actually equivalent.

2. Originally Posted by CoraGB
This problem has left me scratching my head:
The modern (and Archimedean!) meaning of “the series Σ ai (where i goes from 0 to ∞)converges to A" isusually captured by a definition like:

(*)Σ ai (where i goes from 0 to ∞)converges to A if for every ε > 0 there is a K such that for all k K wehave (Σ ai) – A │< ε (where i goes from 0 to k).

Archimedes himself would probably have said something more along the following lines:

(º) Σ ai (where i goes from 0 to ∞)converges to A if both

(1) for every L < A there is a K such that for all k K we have L < (Σ ai) (where i goes from 0 to k),
and

(2) for every U > A there is a Ksuch that for all k Kwe have (Σ ai) < U(where i goes from 0 to k).

Explain, in detail, why these two definitions are actually equivalent.

Attempt - Archimedes definition => Modern Definition
for ε > 0 define L = A- ε and U = A + ε
There is a K such that
all k K we have L < (Σ ai)
There is a K' such that all k K' we have (Σ ai) < U

Let Ko = Max (K,K')
So, for all
k Ko we have L < (Σ ai) < U
i.e A - ε < (Σ ai) < A + ε
i.e.
(Σ ai) – A │< ε

I think other way round can be done on similar lines.
Hence the two definitions are equivatent
Thanks