Question

Use the definition of the limit of a sequence to show that 1 + 1/(2n^2 -n)

convergent sequence with limit 1.

Answer 1

To show that the sequence 1 + 1/(2n^2 - n) converges to the limit 1, we can use the definition of the limit of a sequence. According to the definition, a sequence converges to a limit L if for every positive number ε, there exists a positive integer N such that for all n > N, the terms of the sequence are within ε distance from L. Let's start by simplifying the expression inside the absolute value: |(1 + 1/(2n^2 - n)) - 1| = |1/(2n^2 - n)| = 1/(2n^2 - n). Now, we need to find a value of N that guarantees 1/(2n^2 - n) < ε for all n > N. Let's consider an upper bound for the denominator of the expression: 2n^2 - n > n^2, for all values of n. So, we have 1/(2n^2 - n) < 1/(n^2). Now, set 1/(n^2) < ε and solve for n: n^2 > 1/ε, n > sqrt(1/ε). Therefore, if we choose N = ceil(sqrt(1/ε)), where ceil(x) denotes the smallest integer greater than or equal to x, then for all n > N, 1/(2n^2 - n) < ε. Thus, the sequence converges to the limit 1.

Step-by-step explanation:

To show that the sequence 1 + 1/(2n^2 - n) converges to the limit 1, we can use the definition of the limit of a sequence. According to the definition, a sequence converges to a limit L if for every positive number ε, there exists a positive integer N such that for all n > N, the terms of the sequence are within ε distance from L.

In this case, we want to show that for any ε > 0, there exists N such that for all n > N, |(1 + 1/(2n^2 - n)) - 1| < ε.

Let's start by simplifying the expression inside the absolute value: |(1 + 1/(2n^2 - n)) - 1| = |1/(2n^2 - n)| = 1/(2n^2 - n).

Now, we need to find a value of N that guarantees 1/(2n^2 - n) < ε for all n > N.

Let's consider an upper bound for the denominator of the expression: 2n^2 - n > n^2, for all values of n.

So, we have 1/(2n^2 - n) < 1/(n^2). Now, set 1/(n^2) < ε and solve for n: n^2 > 1/ε, n > sqrt(1/ε).

Therefore, if we choose N = ceil(sqrt(1/ε)), where ceil(x) denotes the smallest integer greater than or equal to x, then for all n > N, 1/(2n^2 - n) < ε. Thus, the sequence converges to the limit 1.