Substitution Not Required

Calculus Level 4

lim ⁡ n → ∞ ∑ r = 1 n r n 2 + n + r = ? \displaystyle \lim_{n \to \infty} \sum_{r=1}^{n} \dfrac{r}{n^2 + n + r} = \ ?

Why no one is trying this problem

0 1 1 3 \dfrac{1}{3} 1 2 \dfrac{1}{2}

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U Z by U Z , 24, Guyana

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2 solutions

If n approach to infinity the denominator will be n 2 n^{2} ,so,it's become the same in all terms.

hence we have

l i m n − > ∞ Σ r = 1 n r n 2 + n + r = l i m n − > ∞ Σ r = 1 n r n 2 = n ( n + 1 ) / 2 n 2 = 1 2 lim_{n -> \infty} \Sigma_{r=1}^{n} \frac{r}{n^{2}+n+r}=lim_{n -> \infty} \Sigma_{r=1}^{n} \frac{r}{n^{2}}=\frac{n(n+1)/2}{n^{2}}=\frac{1}{2}

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Nice solution :) Also, if you want to format an arrow you write: \rightarrow\ in the brackets, and \displaystyle\sum_{r=1}^{n} produces a sum that looks different

Curtis Clement - 6 years, 2 months ago

This solution seems incorrect to me. I have seen this question elsewhere and the solution involves the use of sandwich theorem (which is quite wonderful).

Keshav Tiwari - 5 years, 2 months ago

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I guess you mean this solution.

Let n and r be positive integers. Then

n + r > 0 \displaystyle n+r>0

n 2 + n + r > n 2 \displaystyle { n }^{ 2 }+n+r>{ n }^{ 2 }

r n 2 + n + r < r n 2 \displaystyle \frac { r }{ { n }^{ 2 }+n+r } <\frac { r }{ { n }^{ 2 } }

Also,

r < n + 1 \displaystyle r<n+1

n 2 + n + r < n 2 + 2 n + 1 = ( n + 1 ) 2 \displaystyle { n }^{ 2 }+n+r<{ n }^{ 2 }+2n+1={ (n+1) }^{ 2 }

r ( n + 1 ) 2 < r n 2 + n + r \displaystyle \frac { r }{ { (n+1) }^{ 2 } } <\frac { r }{ { n }^{ 2 }+n+r }

r ( n + 1 ) 2 < r n 2 + n + r < r n 2 \displaystyle \frac { r }{ { (n+1) }^{ 2 } } <\frac { r }{ { n }^{ 2 }+n+r } <\frac { r }{ { n }^{ 2 } }

Hence,

∑ r = 1 n r ( n + 1 ) 2 < ∑ r = 1 n r n 2 + n + r < ∑ r = 1 n r n 2 \displaystyle \sum _{ r=1 }^{ n }{ \frac { r }{ { (n+1) }^{ 2 } } } <\sum _{ r=1 }^{ n }{ \frac { r }{ { n }^{ 2 }+n+r } } <\sum _{ r=1 }^{ n }{ \frac { r }{ { n }^{ 2 } } }

n 2 ( n + 1 ) < S < n + 1 2 n \displaystyle \frac { n }{ { 2(n+1) } } <S<\frac { n+1 }{ 2{ n } }

And so, at the limit

1 2 − < S < 1 2 + \displaystyle { \frac { 1 }{ 2 } }^{ - }<S<{ \frac { 1 }{ 2 } }^{ + }

Or S = 1 2 \displaystyle S={ \frac { 1 }{ 2 } }

คลุง แจ็ค - 5 years, 1 month ago

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Yes. Pretty much it. : ) :)

Keshav Tiwari - 5 years, 1 month ago
Leonel Castillo
Jun 10, 2018

∑ r = 1 n r n 2 + n + r = ∑ r = 1 n 1 n 2 r + n r + 1 = ∑ r = 1 n 1 n + 1 r + 1 n 1 n \sum_{r=1}^n \frac{r}{n^2 + n + r} = \sum_{r=1}^n \frac{1}{\frac{n^2}{r} + \frac{n}{r} + 1} = \sum_{r=1}^n \frac{1}{\frac{n+1}{r} + \frac{1}{n}} \frac{1}{n}

∫ 0 1 x d x = lim ⁡ n → ∞ ∑ r = 1 n r n + 2 1 n ≤ lim ⁡ n → ∞ ∑ r = 1 n 1 n + 1 r + 1 n 1 n ≤ lim ⁡ n → ∞ ∑ r = 1 n 1 n + 1 r 1 n = ∫ 0 1 x d x \int_0^1 x dx = \lim_{n \to \infty} \sum_{r=1}^n \frac{r}{n+2} \frac{1}{n} \leq \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{\frac{n+1}{r} + \frac{1}{n}} \frac{1}{n} \leq \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{\frac{n+1}{r}} \frac{1}{n} = \int_0^1 x dx

∫ 0 1 x d x = 1 2 \int_0^1 x dx = \frac{1}{2}

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