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Limit of a sum (or product) as a definite integral

This type of question, asked 6 times: 2021 Paper 4, 2021 Paper 12, 2021 Paper 18, 2021 Paper 20, 25 Jul 2022, 1 Feb 2023. All 6 questions below, oldest first.

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  1. Question 1

    2021 Paper 4

    70
    lim⁡n→∞[1n+n(n+1)2+n(n+2)2+……+n(2n−1)2]\lim_{n\rightarrow \infty} \left\lbrack \frac{1}{n}+\frac{n}{(n+ 1)^{2}}+\frac{n}{(n+ 2)^{2}}+ \ldots\ldots +\frac{n}{(2n- 1)^{2}} \right\rbrack is equal to:

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  2. Question 2

    2021 Paper 12

    47
    If f:R→Rf:R \rightarrow R is given by f(x)=x+1f(x) = x + 1, then the value of lim⁡n→∞1n[f(0)+f(5n)+f(10n)+…..+f(5(n−1)n)]\lim_{n \rightarrow \infty} \frac{1}{n}\left\lbrack f(0) +f\left( \frac{5}{n} \right)+f\left( \frac{10}{n} \right)+ \ldots.. +f\left( \frac{5(n - 1)}{n} \right) \right\rbrack,is:

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  3. Question 3

    2021 Paper 18

    78
    The value of lim⁡n→∞1n∑r=02n−1n2n2+4r2\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{r= 0}^{2n- 1} \frac{n^{2}}{n^{2}+ 4r^{2}} is:

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  4. Question 4

    2021 Paper 20

    70
    If Un=(1+1n2)(1+22n2)2….(1+n2n2)nU_{n}=\left( 1 +\frac{1}{n^{2}} \right)\left( 1 +\frac{2^{2}}{n^{2}} \right)^{2}\ldots.\left( 1 +\frac{n^{2}}{n^{2}} \right)^{n}, then lim⁡n→∞(Un)−4n2\lim_{n\rightarrow \infty} \left( U_{n} \right)^{\frac{- 4}{n^{2}}} is equal to :

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  5. Question 5

    25 Jul 2022

    157
    lim⁡n→∞12n(11−12n+11−22n+11−32n+…..+11−2n−12n)\lim_{n \rightarrow \infty} \frac{1}{2^{n}}\left( \frac{1}{\sqrt{1 -\frac{1}{2^{n}}}}+\frac{1}{\sqrt{1 -\frac{2}{2^{n}}}}+\frac{1}{\sqrt{1 -\frac{3}{2^{n}}}}+ \ldots.. +\frac{1}{\sqrt{1 -\frac{2^{n}- 1}{2^{n}}}} \right) is equal to

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  6. Question 6

    1 Feb 2023

    73
    lim⁡n→∞[11+n+12+n+13+n+⋯+12n]\lim_{n \rightarrow \infty} \left\lbrack \frac{1}{1 + n}+\frac{1}{2 + n}+\frac{1}{3 + n}+ \cdots +\frac{1}{2n} \right\rbrack is equal to

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