Question: "Prove that limn⇒∞n2n−1=12"

Maximillian Patterson

Answered question

2022-05-01

Prove that

lim_{n \Rightarrow \infty} \frac{n}{2 n - 1} = \frac{1}{2}

Patricia Duffy

Beginner2022-05-02Added 16 answers

Let

S_{n}

be a sequence of real numbers, which is said to be convergent to limit L as n approaches to

\infty

\forall \in > 0 \exists N \in I

such that

| S_{n} - L | < \in, \forall n \geq N

Given

lim_{n \Rightarrow \infty} \frac{n}{2 n - 1} = \frac{1}{2}

Here,

S_{n} = \frac{n}{2 n - 1}, L = \frac{1}{2}

Thus

| S_{n} - L | < \in

| \frac{n}{2 n - 1} - \frac{1}{2} | < \in

| \frac{2 n - 1 (2 n - 1)}{2 (2 n - 1)} | < \in

| \frac{2 n - 2 n + 1}{2 (2 n - 1)} | < \in

| \frac{1}{4 n - 2} | < \in

\frac{1}{4 n - 2} < \in

\frac{1}{4 n} < \in

4 n > \frac{1}{\in}

n > \frac{1}{4 \in} = N

Select

N = [\frac{1}{4 \in}]

\forall \in > 0 \exists N = [\frac{1}{4 \in}]

such that

| \frac{n}{2 n - 1} - \frac{1}{2} | < \in, \forall n \geq N

Hence,

lim_{n \Rightarrow \infty} \frac{n}{2 n - 1} = \frac{1}{2}

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