1. ## sequence 2

General term in a sequence is an=4n+8/(n+5)
What is the sum of greatest lower bound of the sequence and lowest upper bound f the sequence. I got the answer as 5 but it must be 6.

2. ## Re: sequence 2

Originally Posted by kastamonu
General term in a sequence is an=4n+8/(n+5)
What is the sum of greatest lower bound of the sequence and lowest upper bound f the sequence. I got the answer as 5 but it must be 6.

If $n\in\mathbb{Z}^+$ then $2\le\frac{4n+8}{n+5}\le 4$.

3. ## Re: sequence 2

How did you find it?

4. ## Re: sequence 2

VERY EASY...FIND the first term a1=...2 and then the lim of the sequence =4.....then think...it is easy.

5. ## Re: sequence 2

I tried to find the answer by evaluating the fraction. Yes that way is better. Limit is 4.

6. ## Re: sequence 2

4n+8 4n+20-20+8
---- = ----------
n+5 n+5

12
= 4 - ---
n+5 .

Sequence is increasing and lower bound is 2(If we take n=1). If we
find the limit according to infinity upper bound is 4. But I wanted to
find it by using inequality method.

7. ## Re: sequence 2

$\frac{4n+8}{n+5}\le A$ is, since n+ 5 is positive, the same as $4n+ 8\le A(n+ 5)= An+ 5A$. $(A- 4)n\ge 8- 5A$. If we want this to be true for all n, we must have A= 4.

many thanks

9. ## Re: sequence 2

Originally Posted by kastamonu
General term in a sequence is an=4n+8/(n+5)
What is the sum of greatest lower bound of the sequence and lowest upper bound f the sequence. I got the answer as 5 but it must be 6.
\displaystyle \begin{align*} \lim_{n \to \infty} \frac{4n + 8}{n + 5} &= \lim_{n \to \infty} \frac{4n + 20 - 12}{n + 5} \\ &= \lim_{n \to \infty} \left[ \frac{4(n + 5)}{n + 5} - \frac{12}{n + 5} \right] \\ &= \lim_{n \to \infty} \left( 4 - \frac{12}{n + 5} \right) \\ &= 4 - 0 \\ &= 4 \end{align*}

So the least upper bound is 4.