General term in a sequence isa=4n+8/(n+5)_{n}

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.

Printable View

- Feb 25th 2013, 06:47 AMkastamonusequence 2
General term in a sequence is

**a**=4n+8/(n+5)_{n}

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. - Feb 25th 2013, 06:58 AMPlatoRe: sequence 2
- Feb 25th 2013, 07:15 AMkastamonuRe: sequence 2
How did you find it?

- Feb 25th 2013, 07:26 AMMINOANMANRe: sequence 2
VERY EASY...FIND the first term a1=...2 and then the lim of the sequence =4.....then think...it is easy.

- Feb 25th 2013, 07:32 AMkastamonuRe: sequence 2
I tried to find the answer by evaluating the fraction. Yes that way is better. Limit is 4.

- Feb 25th 2013, 10:05 AMkastamonuRe: 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. - Feb 25th 2013, 11:02 AMHallsofIvyRe: sequence 2
$\displaystyle \frac{4n+8}{n+5}\le A$ is, since n+ 5 is positive, the same as $\displaystyle 4n+ 8\le A(n+ 5)= An+ 5A$. $\displaystyle (A- 4)n\ge 8- 5A$. If we want this to be true for all n, we must have A= 4.

- Feb 25th 2013, 10:54 PMkastamonuRe: sequence 2
many thanks

- Feb 25th 2013, 11:39 PMProve ItRe: sequence 2
$\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty} \frac{4n + 8}{n + 5} &= \lim_{n \to \infty} \frac{4n + 20 - 12}{n + 5} \\ &= \lim_{n \to \infty} \left[ \frac{4(n + 5)}{n + 5} - \frac{12}{n + 5} \right] \\ &= \lim_{n \to \infty} \left( 4 - \frac{12}{n + 5} \right) \\ &= 4 - 0 \\ &= 4 \end{align*}$

So the least upper bound is 4.