A problem I saw recently is to compute limn→∞n2((1+1n+1)n+1−(1+1n)n)

Rewrite this asxn=n2(f(1n+1)−f(1n))wheref(x)=(1+x)1xNow, f(x)=exp⁡(1xlog⁡(1+x)) hencef′(x)=g(x)1x2(1+x)f(x)withg(x)=x−(1+x)log⁡(1+x)When x→0, log⁡(1+x)=x−12x2+o(x2))∼−12x2Since f(x)→e when x→0, this yieldsf′(x)sin−12x2⋅1x2⋅e=−12eHencef(1n+1)−f(1n)∼−12e(1n+1−1n)∼12e1n2and finallylimxn=12e

Let consider the sequence: xn=nlog⁡(1+1n) ∼n(1n−1n2+o(1n2)) ∼1−12n+13n2+o(1n2) Then xn+1−xn=(1−12(n+1)+13(n+1)2)−(1−12n+13n2)+o(1n2) =12n(n+1)+13(n+1)2−13n2+o(1n2) =12n2+o(1n2) ∼12n2 consequently n2((1+1n+1)n+1−(1+1n)n)=n2(exn+1−exn) =n2exn(exn+1−xn−1) ∼n2exn(xn+1−xn) ∼exn2 →e2

xn=nlog⁡(1+1n)=1−12n+13n2−14n3+O(1n4) exn=e−e2n+11e24n2−7e16n3+O(1n4) Doing the same and continuing with Taylor series (or long division) xn+1=(n+1)log⁡(1+1n+1)=1−12n+56n2−1712n3+O(1n4) exn+1=e−e2n+23e24n2−89e48n3+O(1n4) exn+1−exn=e2n2−17e12n3+O(1n4) n2(exn+1−exn)=e2−17e12n+O(1n2) which shows th limit and also how it is approached.

Rewrite this asxn=n2(f(1n+1)−f(1n))wheref(x)=(1+x)1xNow, f(x)=exp⁡(1xlog⁡(1+x)) hencef′(x)=g(x)1x2(1+x)f(x)withg(x)=x−(1+x)log⁡(1+x)When x→0, log⁡(1+x)=x−12x2+o(x2))∼−12x2Since f(x)→e when x→0, this yieldsf′(x)sin−12x2⋅1x2⋅e=−12eHencef(1n+1)−f(1n)∼−12e(1n+1−1n)∼12e1n2and finallylimxn=12e

Let consider the sequence: xn=nlog⁡(1+1n) ∼n(1n−1n2+o(1n2)) ∼1−12n+13n2+o(1n2) Then xn+1−xn=(1−12(n+1)+13(n+1)2)−(1−12n+13n2)+o(1n2) =12n(n+1)+13(n+1)2−13n2+o(1n2) =12n2+o(1n2) ∼12n2 consequently n2((1+1n+1)n+1−(1+1n)n)=n2(exn+1−exn) =n2exn(exn+1−xn−1) ∼n2exn(xn+1−xn) ∼exn2 →e2

xn=nlog⁡(1+1n)=1−12n+13n2−14n3+O(1n4) exn=e−e2n+11e24n2−7e16n3+O(1n4) Doing the same and continuing with Taylor series (or long division) xn+1=(n+1)log⁡(1+1n+1)=1−12n+56n2−1712n3+O(1n4) exn+1=e−e2n+23e24n2−89e48n3+O(1n4) exn+1−exn=e2n2−17e12n3+O(1n4) n2(exn+1−exn)=e2−17e12n+O(1n2) which shows th limit and also how it is approached.

We have an answer to "A problem I saw recently is to compute..."

Laney Spears

Answered question

2022-01-21

A problem I saw recently is to compute

lim_{n \to \infty} n^{2} ({(1 + \frac{1}{n + 1})}^{n + 1} - {(1 + \frac{1}{n})}^{n})

Answer & Explanation

fionaluvsyou0x

Beginner2022-01-22Added 11 answers

Rewrite this as

x_{n} = n^{2} (f (\frac{1}{n + 1}) - f (\frac{1}{n}))

where

f (x) = {(1 + x)}^{\frac{1}{x}}

Now,

f (x) = \exp (\frac{1}{x} \log (1 + x))

hence

f^{'} (x) = g (x) \frac{1}{x^{2} (1 + x)} f (x)

with

g (x) = x - (1 + x) \log (1 + x)

When

x \to 0, \log (1 + x) = x - \frac{1}{2} x^{2} + o (x^{2})) \sim - \frac{1}{2} x^{2}

Since

f (x) \to e

when

x \to 0

, this yields

f^{'} (x) \sin - \frac{1}{2} x^{2} \cdot \frac{1}{x^{2}} \cdot e = - \frac{1}{2} e

Hence

f (\frac{1}{n + 1}) - f (\frac{1}{n}) \sim - \frac{1}{2} e (\frac{1}{n + 1} - \frac{1}{n}) \sim \frac{1}{2} e \frac{1}{n^{2}}

and finally

lim x_{n} = \frac{1}{2} e

Kingston Gates

Beginner2022-01-23Added 8 answers

Let consider the sequence:

x_{n} = n \log (1 + \frac{1}{n})

\sim n (\frac{1}{n} - \frac{1}{n^{2}} + o (\frac{1}{n^{2}}))

\sim 1 - \frac{1}{2 n} + \frac{1}{3 n^{2}} + o (\frac{1}{n^{2}})

Then

x_{n + 1} - x_{n} = (1 - \frac{1}{2 (n + 1)} + \frac{1}{3 {(n + 1)}^{2}}) - (1 - \frac{1}{2 n} + \frac{1}{3 n^{2}}) + o (\frac{1}{n^{2}})

= \frac{1}{2 n (n + 1)} + \frac{1}{3 {(n + 1)}^{2}} - \frac{1}{3 n^{2}} + o (\frac{1}{n^{2}})

= \frac{1}{2 n^{2}} + o (\frac{1}{n^{2}})

\sim \frac{1}{2 n^{2}}

consequently

n^{2} ({(1 + \frac{1}{n + 1})}^{n + 1} - {(1 + \frac{1}{n})}^{n}) = n^{2} (e^{x_{n + 1}} - e^{x_{n}})

= n^{2} e^{x_{n}} (e^{x_{n + 1} - x_{n}} - 1)

\sim n^{2} e^{x_{n}} (x_{n + 1} - x_{n})

\sim \frac{e^{x_{n}}}{2}

\to \frac{e}{2}

RizerMix

Expert2022-01-27Added 656 answers

x_{n} = n \log (1 + \frac{1}{n}) = 1 - \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{4 n^{3}} + O (\frac{1}{n^{4}})

e^{x_{n}} = e - \frac{e}{2 n} + \frac{11 e}{24 n^{2}} - \frac{7 e}{16 n^{3}} + O (\frac{1}{n^{4}})

Doing the same and continuing with Taylor series (or long division)

x_{n + 1} = (n + 1) \log (1 + \frac{1}{n + 1}) = 1 - \frac{1}{2 n} + \frac{5}{6 n^{2}} - \frac{17}{12 n^{3}} + O (\frac{1}{n^{4}})

e^{x_{n + 1}} = e - \frac{e}{2 n} + \frac{23 e}{24 n^{2}} - \frac{89 e}{48 n^{3}} + O (\frac{1}{n^{4}})

e^{x_{n + 1}} - e^{x_{n}} = \frac{e}{2 n^{2}} - \frac{17 e}{12 n^{3}} + O (\frac{1}{n^{4}})

n^{2} (e^{x_{n + 1}} - e^{x_{n}}) = \frac{e}{2} - \frac{17 e}{12 n} + O (\frac{1}{n^{2}})

which shows th limit and also how it is approached.

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