This proof splits the sequence x of length n into two subsequences, the first M elements, and the rest.

The property of elements m+1 through n is that |x_{n}| < /2 for n > m

Now given that the sum of the first m+1 elements is finite you can always find some N such that the above some divided by N is also less than /2.

It then composes the magnitude of the average of the n elements of the sequence into the magnitude of the (sum of first m elements + sum of last n-m elements)/n

Using triangle inequality it breaks that up into the

(sum of the magnitudes of first m elements of x)/n + (sum of magnitudes of last (n-m) elements)/n

Now you know by construction that the last (n-m) elements magnitudes are all < /2

So the sum of them divided by n is < (n-m)/n /2 (do you understand this? there are n-m elements in the sum)

Also by construction we know the sum of the first m elements divided by n is < /2.

Combine these and you get the full sum of n magnitudes of x < /2 + (n-m)/n /2 <

So in sum what this does is for a given /2 it finds the element M where the magnitude of x becomes less than /2.

It then takes more elements of the sequence, up to x_{N}so that the sum of the first M elements divided by N < /2. This can be done because the sum of the first M elements is finite so there must be some N big enough.

The (sum of first M elements)/N < /2 because we chose N large enough

The (sum of the last N-M elements)/N < /2 because we chose M large enough.

See if this helps.