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# Suppose that {s_n} is a sequence of positive numbers.?

Prove that if lim (s_n)^2 = s^2 then lim (s_n) = s

### 2 Answers

- kbLv 78 years agoFavorite Answer
Since {s(n)} is a sequence of positive numbers, its limit can not be negative (ruling out -s, assuming that s is non-negative).

We know that {(s(n))^2} is bounded (as it converges):

So, 0 < (s(n))^2 < M for some M > 0 (and for all n)

==> 0 < s(n) < √M for all n.

Now, let ε > 0 be given.

Since {(s(n))^2} converges to s^2, there exists a positive integer N such that

|(s(n))^2 - s^2| < ε/(s + √M) for all n > N.

So, |(s(n))^2 - s^2|

= |s(n) - s| * (s(n) + s)

< |s(n) - s| * ε/(s + √M)

= ε for all n > N.

Hence, {s(n)} converges to s.

I hope this helps!

- RaffaeleLv 78 years ago
This is a particular case of the more general property

Let f be any continuous function,

if an → a then f(an) → f(a)

in your question f(x) = √x

http://en.wikipedia.org/wiki/Limit_of_a_sequence#P...

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if (Sn)² → S² then Sn → S

suppose S > 0

let ε> 0 be any positive real number such that

ε < S.............(*)

ε² < S², there exists an index N such that

if n>N then |S² - (Sn)²| < ε²

S² - ε² < (Sn)² < S² + ε²

S² + ε² < (S + ε)² = S² + ε² + 2εS

therefore

√(S² + ε²) < S + ε,

for positive S and ε

√(S² - ε²) > S - ε

square both sides

S² - ε² > S² + ε² - 2εS

- 2 ε² > - 2εS

ε < S, which is true for (*)

from

S² - ε² < (Sn)² < S² + ε²

taking square root of all terms

S - ε < √(S² - ε²) < Sn < √(S² + ε²) < S + ε,

S - ε < Sn < S + ε,

|S - Sn| < ε

QDE