Show that if Σ Ak from 1 to ∞ is convergent, then ∀ ε > 0 ∃ N ∈ the naturals such that m ≥ N ==> | Σ Ak from m?
Show that if Σ Ak from 1 to ∞ is convergent,
then ∀ ε > 0 ∃ N ∈ the naturals such that m ≥ N ==>
|Σ Ak| from m to ∞ < ε
Note that we know that Σ Ak from m to ∞ = (Σ Ak from 1 to ∞) - (Σ Ak from 1 to m-1)
now I think I can apply the definition of a limit but I am struggling to do this.
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The main idea is to recall that
Σ(k=1 to ∞) A(k) = lim(n→∞) Σ(k=1 to n) A(k).
That is, S = Σ(k=1 to ∞) A(k) is the limit of the sequence of partial sums {S(n)}, where
S(n) = Σ(k=1 to n) A(k).
Since {S(n)} converges to S by hypothesis, the definition of convergence yields:
∀ε > 0 ∃ positive integer N such that |S - S(m-1)| < ε for all m ≥ N.
Now, rewrite this in terms of sums:
∀ε > 0 ∃ positive integer N such that |Σ(k=1 to ∞) A(k) - Σ(k=1 to m-1) A(k)| < ε for all m ≥ N.
==> ∀ε > 0 ∃ positive integer N such that |Σ(k=m to ∞) A(k)| < ε for all m ≥ N.
I hope this helps!