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Establish the given limit relations

by RoRi

Use the previous exercise (Section 10.4, Exercise #34) to establish each of the following limits.

  1. \displaystyle{\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \frac{k}{n} \right)^2 = \frac{1}{3}}.
  2. \displaystyle{ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k} = \log 2}.
  3. \displaystyle{ \lim_{n \to \infty} \sum_{k=1}^n \frac{n}{n^2 + k^2} = \frac{\pi}{4}}.
  4. \displaystyle{ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{n^2 + k^2}} = \log \left( 1 + \sqrt{2} \right)}.
  5. \displaystyle{ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \sin \frac{k \pi}{n} = \frac{2}{\pi}}.
  6. \displaystyle{ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \sin^2 \frac{k \pi}{n} = \frac{1}{2}}.
  1. Let f(x) = x^2, then from Exercise #34 we know

        \[ 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n} \qquad \text{for all } n, \]

    where

        \[ s_n = \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right). \]

    Thus,

        \[ \lim_{n \to \infty} \int_0^1 f(x) \, dx - \frac{1}{n} \sum_{k=0}^{n-1} \left( \frac{k}{n} \right)^2 = 0 \]

    (since \lim_{n \to \infty} \frac{f(1) - f(0)}{n} = 0 and then use the squeeze theorem). So,

        \begin{align*} \implies && \int_0^1 f(x) \, dx &= \frac{1}{n} \sum_{k=0}^{n-1} \left( \frac{k}{n} \right)^2 \\[9pt] \implies && \frac{1}{n} \sum_{k=0}^{n-1} \left( \frac{k}{n} \right)^2 &= \int_0^1 x^2 \, dx \\[9pt] &&&= \frac{1}{3}. \end{align*}

  2. Let

        \[ f(x) = \frac{1}{1+x} \qquad \implies \qquad f \left( \frac{k}{n} \right) = \frac{n}{n+k}. \]

    So,

        \[ \frac{1}{n} \sum_{k=1}^n f \left( \frac{k}{n} \right) = \sum_{k=1}^n \frac{1}{n+k}. \]

    Thus,

        \begin{align*} && 0 \leq \int_0^1 f(x) \, dx - \sum_{k=1}^n \frac{1}{n+k} &\leq \frac{f(1) - f(0)}{n} \\[9pt] \implies && \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - \sum_{k=1}^n \frac{1}{n+k} \right) &= 0 \\[9pt] \implies && \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k} &= \int_0^1 \frac{1}{1+x} \, dx \\[9pt] &&&= \log (1+x) \Bigr \rvert_0^1 \\[9pt] &&&= \log 2. \end{align*}

  3. Let

        \[ f(x) = \frac{1}{1+x^2} \qquad \implies \qquad f\left( \frac{k}{n} \right) = \frac{1}{1 + \left( \frac{k}{n} \right)^2} = \frac{n^2}{n^2+k^2}. \]

    Thus,

        \[ \frac{1}{n} \sum_){k=1}^n f \left( \frac{k}{n} \right) = \sum_{k=1}^n \frac{n}{n^2 + k^2}. \]

    Therefore,

        \begin{align*} && 0 \leq \left( \int_0^1 f(x) \, dx - \sum_{k=1}^n \frac{n}{n^2 + k^2} \right) &\leq \frac{f(1) - f(0)}{n} \\[9pt] \implies && \lim_{n \to \infty} \sum_{k=1}^n \frac{n}{n^2 + k^2} &= \int_0^1 \frac{1}{1+x^2} \, dx \\[9pt] &&&= \arctan x \Bigr \rvert_0^1 \\[9pt] &&&= \frac{\pi}{4}. \end{align*}

  4. Let

        \[ f(x) = \frac{1}{\sqrt{1+x^2}} \quad \implies \quad f\left( \frac{n}{k} \right) = \frac{1}{\sqrt{1+ \left( \frac{k}{n} \right)^2}} = \frac{n}{\sqrt{n^2 + k^2}}. \]

    Thus,

        \[ \frac{1}{n} \sum_{k=1}^n f \left( \frac{n}{k} \right) = \sum_{k=1}^n \frac{n}{\sqrt{n^2+k^2}}. \]

    So,

        \begin{align*} && 0 \geq -\int_0^1 f(x) \, dx + \sum_{k=1}^n \frac{n}{\sqrt{n^2+k^2}} &\geq \frac{f(0)-f(1)}{n} \\[9pt] \implies && \lim_{n \to \infty} \sum_{k=1}^n \frac{n}{\sqrt{n^2+k^2}} &= \int_0^1 f(x) \, dx \\[9pt] &&&= \int_0^1 \frac{1}{\sqrt{1+x^2}} \, dx \\[9pt] &&&= \int_0^1 \frac{x+\sqrt{1+x^2}}{(\sqrt{1+x^2})(x + \sqrt{1+x^2})} \, dx \\[9pt] &&&= \int_0^1 \frac{x+\sqrt{1+x^2}}{x\sqrt{1+x^2} + 1 + x^2} \, dx \\[9pt] &&&= \int_0^1 \left( \frac{1}{x+\sqrt{1+x^2}} \right) \left( \frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right) \, dx \\[9pt] &&&= \log \left( x + \sqrt{1+x^2} \right) \Bigr \rvert_0^1 \\[9pt] &&&= \log \left( 1 + \sqrt{2} \right). \end{align*}

  5. Let

        \[ f(x) = \sin (\pi x) \qquad \implies \qquad f \left( \frac{k}{n} \right) = \sin \frac{k \pi}{n}. \]

    Thus,

        \[ \frac{1}{n} \sum_{k=1}^n f \left( \frac{k}{n} \right) = \sum_{k=1}^n \frac{1}{n} \sin \frac{k \pi}{n}. \]

    Therefore,

        \begin{align*} && \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \sin \frac{k \pi}{n} &= \int_0^1 \sin (\pi x) \, dx \\[9pt] \implies && \lim_{n \to \infty} \sum_{k=1}^n \sin \frac{k \pi}{n} &= \frac{1}{\pi}\left( - \cos (\pi x) \Bigr \rvert_0^1 \right) \\[9pt] &&&= \frac{2}{\pi}. \end{align*}

  6. Let

        \[ f(x) = \sin^2 (\pi x) \qquad \implies \qquad f \left( \frac{k}{n} \right) = \sin^2 \frac{k \pi}{n}. \]

    Thus,

        \[ 0 \geq \sum_{k=1}^n \frac{1}{n} \sin^2 \frac{k \pi}{n} - \int_0^1 f(x) \, dx \geq \frac{f(0) - f(1)}{n}. \]

    Therefore,

        \begin{align*} && \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \sin^2 \frac{k \pi}{n} &= \int_0^1 \sin^2 (\pi x) \, dx \\[9pt] &&& =\frac{1}{2} \int_0^1 (1 - \cos (2 \pi x)) \, dx \\[9pt] &&& = \frac{1}{2} \left( -\frac{1}{4 \pi} \sin (2 \pi x) \Bigr \rvert_0^1 \right) \\[9pt] &&& = \frac{1}{2}. \end{align*}

3 comments

  1. Mohammad Azad says:

    First, most of these functions are not increasing but this can be fixed by multiplying by -1, however there is big problem with (e) since it’s not monotonic… but I managed to prove it by substituting 2n instead of n and then breaking the sum into two parts one is usual since sin(xpi/2) is increasing on [0,1] and the other part you just change the indexing and you get a cosine which can be dealt with as usual.

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