Question

Given a sequence f:Z+→R of real numbers, we say that the sequence converges to a real number L if for all ε>0, there exists a positive integer N such that for any positive integer n, if n≥N, then |f(n)-L|<ε. Prove that if a sequence f converges to L and a sequence g converges to M, then the sequence f+g converges to L+M. (You may use the triangle inequality: |a+b|≤|a|+|b| for any real numbers a and b.)

Answer 1

See proof below

Explanation:

If the sequence f converges to L and the sequence g converges to M, then for any
`\bf \epsilon >0` there exists
`\bf N-1` and
`\bf N_2` such that

`\bf |f(n)-L|<\epsilon/2 \;for\;n>N_1\\|g(n)-M|<\epsilon/2 \;for\;n>N_2`

But then, by the triangle inequality

`\bf |(f(n)+g(n))-(L+M)|=|(f(n)-L)+(g(n)-M)|\leq\\|f(n)-L|+|g(n)-M|<\epsilon/2+\epsilon/2=\epsilon`

for
`\bf N>Max\{N_1,N_2\}`

Hence

f+g converges to L+M