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Задача 1.1. Сходиттся ли пос́лсдоватсльность \[ x_{n}=\left(1, \frac{1}{2}, \fr...
Jun 28, 2024
Solution by Steps
step 1
We need to determine if the sequence xn=(1,12,122,…,12n−1,0,0,…)x_n = \left(1, \frac{1}{2}, \frac{1}{2^2}, \ldots, \frac{1}{2^{n-1}}, 0, 0, \ldots\right) converges in the space l1l_1
step 2
The space l1l_1 consists of all sequences whose series of absolute values is convergent. Therefore, we need to check if the series ∑n=1∞∣xn∣\sum_{n=1}^{\infty} \left| x_n \right| converges
step 3
The given sequence can be written as ∑n=1∞12n−1\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}
step 4
This is a geometric series with the first term a=1a = 1 and common ratio r=12r = \frac{1}{2}
step 5
The sum of an infinite geometric series is given by a1−r\frac{a}{1 - r}, where |r| < 1
step 6
Substituting the values, we get ∑n=1∞12n−1=11−12=2\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = \frac{1}{1 - \frac{1}{2}} = 2
step 7
Since the series converges to 2, the sequence xnx_n converges in the space l1l_1
step 8
The limit of the sequence as n→∞n \to \infty is 0, as lim⁡n→∞(12)n=0\lim_{n \to \infty} \left( \frac{1}{2} \right)^n = 0
Answer
The sequence xnx_n converges in the space l1l_1 and the limit of the sequence is 0.
Key Concept
Convergence of a sequence in l1l_1 space
Explanation
A sequence converges in the l1l_1 space if the series of its absolute values is convergent. In this case, the given sequence forms a geometric series that converges to 2, hence it converges in l1l_1.
Solution by Steps
step 1
We need to determine if the sequence xn(t)=nt21+n+t x_{n}(t) = \frac{n t^{2}}{1+n+t} converges in the space L1(0,1) L^{1}(0,1)
step 2
To check for convergence in L1(0,1) L^{1}(0,1) , we need to evaluate the limit as n n approaches infinity of the integral from 0 to 1 of xn(t) x_{n}(t) . This is given by: lim⁡n→∞∫01nt21+n+t dt \lim_{n \to \infty} \int_{0}^{1} \frac{n t^{2}}{1+n+t} \, dt
step 3
We simplify the integrand: nt21+n+t≈nt2n=t2 as n→∞ \frac{n t^{2}}{1+n+t} \approx \frac{n t^{2}}{n} = t^{2} \text{ as } n \to \infty Thus, the integral becomes: lim⁡n→∞∫01t2 dt \lim_{n \to \infty} \int_{0}^{1} t^{2} \, dt
step 4
Evaluate the integral: ∫01t2 dt=[t33]01=13 \int_{0}^{1} t^{2} \, dt = \left[ \frac{t^{3}}{3} \right]_{0}^{1} = \frac{1}{3}
step 5
Since the limit of the integral is finite, the sequence xn(t) x_{n}(t) converges in L1(0,1) L^{1}(0,1)
Answer
The sequence xn(t)=nt21+n+t x_{n}(t) = \frac{n t^{2}}{1+n+t} converges in the space L1(0,1) L^{1}(0,1) and the limit of the integral is 13 \frac{1}{3} .
Key Concept
Convergence in L1(0,1) L^{1}(0,1)
Explanation
To determine convergence in L1(0,1) L^{1}(0,1) , we evaluate the limit of the integral of the sequence's terms. If the limit is finite, the sequence converges in L1(0,1) L^{1}(0,1) .
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