I would like to prove (or disprove) the following limit:limn→∞cn∑j=1n(1−bcn)2(n−j)=12b(1−e−2bc)for b,c&gt;0

Take x=bcn. Then n=bcx and x→0+ as n→+∞. Moreover,x(x−1)2−1→−12(1−x)2bcx→e−2bcas x→0+. Thereforecn(1−bcn)2n−1(1−bcn)2−1=x(x−1)2−1⋅1b⋅[(1−x)2bcx−1]→−12⋅1b⋅[e−2bc−1]

As an alternative we havecn(1−bcn)2n−1(1−bcn)2−1=cn(1−bcn)2n−1(1−bcn+1)(1−bcn−1)=−nbccn(1−bcn)2n−1(2−bcn)→−12b(e−2bc−1)

Take x=bcn. Then n=bcx and x→0+ as n→+∞. Moreover,x(x−1)2−1→−12(1−x)2bcx→e−2bcas x→0+. Thereforecn(1−bcn)2n−1(1−bcn)2−1=x(x−1)2−1⋅1b⋅[(1−x)2bcx−1]→−12⋅1b⋅[e−2bc−1]

As an alternative we havecn(1−bcn)2n−1(1−bcn)2−1=cn(1−bcn)2n−1(1−bcn+1)(1−bcn−1)=−nbccn(1−bcn)2n−1(2−bcn)→−12b(e−2bc−1)

Best solutions to - "I would like to prove (or disprove) the..."

Ivor Schofield

Answered question

2022-02-25

I would like to prove (or disprove) the following limit:

lim_{n \to \infty} \frac{c}{n} \sum_{j = 1}^{n} {(1 - \frac{b c}{n})}^{2 (n - j)} = \frac{1}{2 b} (1 - e^{- 2 b c})

for

b, c > 0

Answer & Explanation

aceptanteppt

Beginner2022-02-26Added 5 answers

Take

x = \frac{b c}{n}

. Then

n = \frac{b c}{x}

and

x \to 0^{+}

n \to + \infty

. Moreover,

\frac{x}{{(x - 1)}^{2} - 1} \to - \frac{1}{2}

{(1 - x)}^{\frac{2 b c}{x}} \to e^{- 2 b c}

x \to 0^{+}

. Therefore

\frac{c}{n} \frac{{(1 - \frac{b c}{n})}^{2 n} - 1}{{(1 - \frac{b c}{n})}^{2} - 1} = \frac{x}{{(x - 1)}^{2} - 1} \cdot \frac{1}{b} \cdot [{(1 - x)}^{\frac{2 b c}{x}} - 1]

\to - \frac{1}{2} \cdot \frac{1}{b} \cdot [e^{- 2 b c} - 1]

Jett Brooks

Beginner2022-02-27Added 7 answers

As an alternative we have

\frac{c}{n} \frac{{(1 - \frac{b c}{n})}^{2 n} - 1}{{(1 - \frac{b c}{n})}^{2} - 1} = \frac{c}{n} \frac{{(1 - \frac{b c}{n})}^{2 n} - 1}{(1 - \frac{b c}{n} + 1) (1 - \frac{b c}{n} - 1)}

= - \frac{n}{b c} \frac{c}{n} \frac{{(1 - \frac{b c}{n})}^{2 n} - 1}{(2 - \frac{b c}{n})} \to - \frac{1}{2 b} (e^{- 2 b c} - 1)

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