Answer :
Answer:
The sequence converges to 5/9.
Step-by-step explanation:
Suppose we have a sequence [tex]a_{n}[/tex]
Limit test:
[tex]L = \lim_{n \to \infty} a_{n}[/tex]
If L is a constant number, the sequence converges to L.
If L = 0, the test is inconclusive.
Otherwise, the sequence diverges.
In this question:
[tex]a_{n} = \frac{5n}{1 + 9n}[/tex]
Test:
[tex]L = \lim_{n \to \infty} (\frac{5n}{1 + 9n})[/tex]
Infinity limit, we consider the term with the highest degree in the numerator and in the denominator.
[tex]L = \lim_{n \to \infty} (\frac{5n}{1 + 9n}) = \lim_{n \to \infty} \frac{5n}{9n} = \lim_{n \to \infty} \frac{5}{9} = \frac{5}{9}[/tex]
The sequence converges to 5/9.
You can check whether the sequence is decreasing or not in this case and use them with the limit.
The answers are:
a) The sequence converges. The limit is 5/9
b) The sequence converges. The limit is 0
How to find if a sequence is converging or not?
There are many methods to find that. But one reverse test to be sure that the given sequence is not converging is that the limit when n tends to infinity, won't come as a finite point.
How to test the convergence for first sequence given?
[tex]a_n = \dfrac{5n}{1+9n}[/tex]
We have this sequence upper bounded by: [tex]\dfrac{5n}{9n} = \dfrac{5}{9}[/tex] (assuming n starts from 1) since the less the denominator, the bigger the term is (since 9n is positive and 1 is positive so 1 + 9n > 9n )
The sequence is lower bounded by [tex]\dfrac{5n}{n + 9n} = \dfrac{5n}{10n} = \dfrac{5}{10} = \dfrac{1}{2}[/tex]
Checking if the sequence is monotonic:
[tex]a_{n+1} - a_n = \dfrac{5(n+1)}{1+9(n+1)} - \dfrac{5n}{1+9n} = \dfrac{(5n + 5)(1+9n) - (5n)(10 + 9n)}{(10 + 9n)(1+ 9n)} \\\\=a_{n+1} - a_n = \dfrac{5}{(10+9n)(1+9n)} > 0[/tex]
Thus the sequence is monotonically increasing sequence.
Thus, since the sequence is bounded and monotonic, thus it is converging.
The limit of this sequence is found by:
[tex]lim_{n \rightarrow \infty}\dfrac{5n}{1+9n} = lim_{n \rightarrow \infty}\dfrac{5}{1/n+9} = \dfrac{5}{9}[/tex]
Thus, the limit of this sequence is 5/9
Checking the convergence of second given sequence:
Since the given sequence is always bigger than 0, thus it is lower bounded by 0.
Now the given sequence is monotonic decreasing since each term is less than or equal to previous term. Thus, due to it being monotonic decreasing and being bounded (and all terms are smaller or equal to 1, thus upper bounded too), we have this sequence as converging.
The limit of this sequence is taken by:
[tex]a_n = \dfrac{1}{mod(n, 2)}\\\\\rm As \: n \rightarrow \infty, \text{ we have mod(n,2)} \rightarrow \infty\\Thus, a_{\infty}= 0[/tex]
Thus, the given sequence converges to 0.
Learn more about convergence of sequence here:
https://brainly.com/question/20945705