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# Prove that the following sequence is bounded...?

a_n = 2^n / [2^n + 3^n]

Prove that it's bounded by - 2/3 (lower) and 2/3 (upper).

Start with a_n and work to your answer. Show your work. Thank you! :)

### 1 Answer

- 1 decade agoFavorite Answer
Use induction for both.

Pf bounded above by 2/3:

Base: (n=1) 2^1/ (2^1 + 3^1) = 2/5 < 2/3. done

Induction Hypothesis: 2^k / (2^k + 3^k) < 2/3.

Show k => k+1:

2^(k+1)/ [2^(k+1) + 3^(k+1)]

= 2(2^k) / [2(2^k) + 3(3^k)] ............write in terms of k....

< 2(2^k) / [2(2^k) + 2(3^k)]..............change 3 to a 2 in denominator......

= 2^k / [2^k + 3^k] ...........................pull out 2/2 = 1........

< 2/3 .................................................by induction hypothesis

QED

Pf bounded below by -2/3:

Base: (n=1) 2/5 > -2/3. done.

Induction Hypothesis: 2^k / (2^k + 3^k) > -2/3

Show k => k+1:

2^(k+1) / [2^(k+1) + 3^(k+1)]

= 2(2^k) / [2(2^k) + 3(3^k)] ................write in terms of k.....

> 2^k / [ 3(2^k) + 3 (3^k) ] .........remove 2 from numerator and change 2 to 3 in denominator.....

= (1/3) [2^k /(2^k + 3^k)] ........pull 1/3 out......

> -2/9 ......................by induction hypothesis

> -2/3

QED

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