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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.)

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rodolforodriguezr

Answer:

See proof below

Step-by-step explanation:

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

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

But then, by the triangle inequality

[tex] \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[/tex]

for [tex] \bf N>Max\{N_1,N_2\}[/tex]

Hence

f+g converges to L+M

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