Show that a sequence {sn} coverages to a limit l if and only if the sequence {sn-l} coverages to zero.

Step-by-step explanation:

To prove this we can use the definition of a sequence converging to its limit, in terms of epsilon:

The sequence converges to

if and only if

for every there exists such that

if and only if

for every there exists such that

if and only if

the sequence converges to 0.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.