## osanseviero Group Title Determine if the next progression have a limit (demostrate it) and determine it 8 months ago 8 months ago

1. osanseviero Group Title

$\left( -1^{n} \right)\left( \frac{ 5n+4 }{ 2n } \right)$

2. osanseviero Group Title

3. osanseviero Group Title

:/

4. tkhunny Group Title

As n increases in the positive direction, $$\dfrac{5n+4}{2n}$$ approaches $$\dfrac{5}{2}$$. Follow with your mind as n increases. The little 4 on the end of the numerator becomes less and less significant. The terms do NOT approach zero.

5. osanseviero Group Title

oh...I think I see it now...but how can i demostrate if it is it's limit?

6. osanseviero Group Title

but there is also the -1^n

7. osanseviero Group Title

So it approaches 5/2 and -5/2 ?

8. tkhunny Group Title

The typical demonstration is a division by n. For n > 0, $$\dfrac{5n+4}{2n} = \dfrac{5 + \dfrac{4}{n}}{2}$$. In this form, it is relatively obvious that the limit it 5/2 as n increases. The FIRST criterion for convergence is terms that approach ZERO. Nothing else will do. These terms do not approach zero, therefore, we do not care about the alternating sign. If the terms approach zero, THEN we'll worry about the sign.

9. osanseviero Group Title

what I mean is that there is a (-1^n) multiplying all of that...so -5/2 is also a limit

10. tkhunny Group Title

No, this is not a limit. Limits come alone, not in pairs. The terms, without the sign, approach 5/2. I may have stated that carelessly, before. The actual terms, including the sign, do not have a limit. They is oscillating.

11. osanseviero Group Title

Oh...okk

12. osanseviero Group Title

So for this there isnt a limit, neither $\frac{ 1 }{ 2 },2^{2}, \frac{ 1 }{ 2^{3} }$

13. tkhunny Group Title