I am maybe overthinking this, but what is the lim as n-> infinity of |(n+1)/(n+2)|
?
I am trying to use the ratio test to find interval of convergence for the infiinite series (n/n+1)((-2x)^(n-1))
so i did the limit of |(n+1)/(n+2)| but i am wondering if the limit is simply =1, or if it is infinity/infinity so use l'hopital's rule???
maybe i am overthinking it and the limit is just 1.
thank you for your assistance
the limit of your problem is 1
as n --> infinity , both numerators and denominator are practiacally the same, with the denominator being one less than the numerator.
e.g. if n = 1 000 000
then we have 1 000 001/1 000 002 , pretty close to 1 wouldn't you say?
since (n+1)/(n+2) = 1 - 1/(n+2) it is clear that as n grows, the value approaches 1.
Yes, L'Hopital's Rule also works, since you have infinity/infinity. taking derivatives gives 1/1 as the limit.
To find the limit as n approaches infinity of |(n+1)/(n+2)|, we can simplify the expression by dividing both the numerator and denominator by n. This gives us:
|(1+1/n)/(1+2/n)|
As n approaches infinity, both 1/n and 2/n approach zero. Therefore, the limit can be simplified further:
|1/1|
And since absolute value is always positive, the final answer is 1.
Regarding the ratio test to find the interval of convergence for the infinite series (n/(n+1))((-2x)^(n-1)), you are correct in taking the limit of |(n+1)/(n+2)|. In this case, you don't need to apply L'Hopital's rule as the limit is already straightforward.
The limit is 1, meaning that the ratio of consecutive terms in the series approaches 1. To determine the interval of convergence, it is necessary to set up an inequality with the ratio less than 1:
|((n+1)/(n+2))| < 1
Since the absolute value of a positive number is always less than 1, the inequality is true for all values of n. Therefore, the interval of convergence for the infinite series is the entire real number line (-∞, +∞).
So, in this case, you were not overthinking it, and the limit is indeed 1.