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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*}

Find the Taylor polynomial of log ((1+x)/(1-x))1/2

by RoRi

Show that

    \[ T_{2n+1} \left( \log \sqrt{ \frac{1+x}{1-x}} \right) = \sum_{k=0}^n \frac{x^{2k+1}}{2k+1}. \]

First, we have

    \[ f(x) = \log \sqrt{ \frac{1+x}{1-x}} = \frac{1}{2} ( \log (1+x) - \log(1-x)). \]

We know from the previous exercise (Section 7.4, Exercise #6) that

    \[ T_n (\log(1+x)) = \sum_{k=1}^n \frac{(-1)^{k+1}x^k}{k}. \]

We also know (from the example on page 277) that

    \[ T_{n+1} (-\log(1-x)) = \sum_{k=1}^{n+1} \frac{x^k}{k}. \]

Therefore, using Theorem 7.2 (a), the linearity property of T_n we have

    \begin{align*} T_{2n+1} f(x) &= \frac{1}{2} T_{2n+1} \log(1+x) + \frac{1}{2} T_{2n+1}(-\log(1-x)) \\[9pt] &= \frac{1}{2} \left( \sum_{k=1}^{2n+1} \frac{(-1)^{k+1}x^k}{k} + \sum_{k=1}^{2n+1} \frac{x^k}{k} \right) \\[9pt] &= \frac{1}{2} \left( \sum_{k=1}^{2n+1} \frac{(-1)^{k+1} x^k + x^k}{k} \right) \\[9pt] &= \frac{1}{2} \sum_{k=1}^{2n+1} \frac{x^k ((-1)^{k+1} + 1)}{k}. \end{align*}

However, (-1)^{k+1} + 1 = 0 if k is even and (-1)^{k+1} + 1 = 2 if k is odd. Therefore we can sum over just the odd values of k. Let k = 2j-1 and we have

    \begin{align*} T_{2n+1} \sqrt{\frac{\log (1+x)}{\log(1-x)}} &= \frac{1}{2} \sum_{k=1}^{2n+1} \frac{x^k ((-1)^{k+1} + 1)}{k} \\[9pt] &= \frac{1}{2} \sum_{j=1}^{n+1} \frac{x^{2j-1} (2)}{2j-1} \\[9pt] &= \sum_{j=1}^{n+1} \frac{x^{2j-1}}{2j-1} \\[9pt] &= \sum_{j=0}^n \frac{x^{2j+1}}{2j+1} \\[9pt] &= \sum_{k=0}^n \frac{x^{2k+1}}{2k+1}. \end{align*}

Where we have renamed the index of summation in the final step so that the sum is over k as in the book.

Prove inequalities of the logarithm with respect to some series

by RoRi

Consider a partition P = \{ a_0, a_1, a_2, \ldots, a_n \} of the interval [1,x] for some x > 1.

  1. Find step functions that are constant on the open subintervals of P and integrate to derive the inequalities:

        \[ \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) < \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right). \]

  2. Give a geometric interpretation of the inequalities in part (a).
  3. Find a particular partition P (i.e., choose particular values for a_0, a_1, \ldots, a_n) to establish the following inequalities for n > 1,

        \[ \sum_{k=2}^n \frac{1}{k} < \log n < \sum_{k=1}^{n-1} \frac{1}{k}. \]

  1. Proof. We define step function s and t by

        \begin{align*} s(x) &= \frac{1}{a_k} & \text{for } x &\in [a_{k-1}, a_k ) \\ t(x) &= \frac{1}{a_{k-1}} & \text{for } x &\in [a_{k-1}, a_k ). \end{align*}

    Since \frac{1}{x} is strictly decreasing on \mathbb{R}_{>0}, we have

        \[ s(x) < \frac{1}{x} < t(x) \qquad \text{for all } x \in \mathbb{R}_{>0}. \]

    Therefore, using the definition of the integral of a step function as a sum,

        \begin{align*} && \int_1^x s(u) \, du &< \int_1^x \frac{1}{u} \, du < \int_1^x t(u) \, du \\ \implies && \sum_{k=1} s_k (a_k - a_{k-1} ) &< \log x < \sum_{k=1}^n t_k (a_k - a_{k-1}) \\ \implies && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right). \qquad \blacksquare \end{align*}

  2. Geometrically, these inequalities say that the area under the curve \frac{1}{x} lies between the step functions that take on the values \frac{1}{a_k} and \frac{1}{a_{k-1}} for each x \in [ a_{k-1}, a_k ).
  3. Proof. To establish these inequalities we pick the partition,

        \[ P = \{ 1, 2, 3, \ldots, n \} \qquad \text{for } n > 1. \]

    Then, applying part (a) we have

        \begin{align*} && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right) \\ \implies && \sum_{k=1}^n \frac{1}{a_k} &< \log n < \sum_{k=1}^n \frac{1}{a_{k-1}} &(\text{since } a_k - a_{k-1} = 1) \\ \implies && \sum_{k=2}^n \frac{1}{k} &< \log n < \sum_{k=1}^{n-1} \frac{1}{k}. \end{align*}

    The final line follows since a_0 = 1, a_1 = 2, \ldots so the sum on the left starts with \frac{1}{2} and the sum on the right only runs to \frac{1}{n-1}. These were the inequalities requested. \qquad \blacksquare

Determine formulas for given sums

by RoRi

Consider the formula

    \[ 1 + x + x^2 + \cdots + x^n = \frac{x^{n+1} - 1}{x-1} \qquad (x \neq 1). \]

By differentiating, determine formulas for the following:

  1. 1 + 2x + 3x^2 + \cdots + nx^{n-1}.
  2. 1^2 x + 2^2 x^2 + 3^2 x^3 + \cdots + n^2 x^n.
  1. We observe that

        \[ \left(1 + x + x^2 + \cdots + x^n\right)' = 1 + 2x + 3x^2 + \cdots + nx^{n-1}. \]

    Thus, using the given formula we have,

        \begin{align*} 1 + 2x + 3x^2 + \cdots + nx^{n-1} &= \left( \frac{x^{n+1} - 1}{x-1} \right)' \\ &= \frac{(n+1)x^n (x-1) - x^{n+1} + 1}{(x-1)^2} \\ &= \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2}. \end{align*}

  2. (Note: There’s an alternate solution to this in the comments that is more direct than this one. Definitely worth taking a look at that alternative.)
    Taking derivatives of both sides of the equation we derived in part (a),

        \begin{align*} &&\left(1+2x + 3x^2 + \cdots + nx^{n-1}\right)' &= \left( \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \right)' \\[10pt] \implies && 2 + 6x + \cdots + n(n-1)x^{n-2}& \\ &&& \hspace{-4.2cm} = \frac{(x-1)^2 (n(n+1)x^n - n(n+1)x^{n-1}) - (2x-2)(nx^{n+1} - (n+1)x^n + 1)}{(x-1)^4} \\[10pt] \implies && (1 + 4x + 9x^2 + \cdots + (n-1)^2& x^{n-2}) + (1 + 2x + 3x^2 + \cdots + (n-1)x^{n-2}) \\ &&& \hspace{-4.2cm}= \frac{(x-1)(n(n+1)x^n - n(n+1)x^{n-1}) - 2(nx^{n+1} - (n+1)x^n + 1)}{(x-1)^3}\\[10pt] \implies && \left( \frac{1}{x} \right) (1^2 x + 2^2 x^2 + \cdots + (n-1)^2 &x^{n-1}) + (1 + 2x + \cdots + (n-1)x^{n-2}) \\ &&& \hspace{-4.2cm}= \frac{(n^2-n)x^{n+1} - 2(n^2-1)x^n + n(n+1)x^{n-1} - 2}{(x-1)^3}. \end{align*}

    Then, looking at the second term in the sum on the left, we have,

        \begin{align*} 1+ 2x + \cdots + (n-1)x^{n-2} &= (1 + x + \cdots + x^n)' - nx^{n-1} \\ &= \left( \frac{x^{n+1}-1}{x-1} \right)' - nx^{n-1} \\ &= \frac{nx^{n+1} - (n+1)x + 1}{(x-1)^2} - nx^{n-1} \\ &= \frac{(n+1)x^n - nx^{n-1} + 1}{(x-1)^2}. \end{align*}

    Thus, plugging this back into the expression above,

        \begin{align*} && \frac{1}{x} &\left(1^2 x + 2^2 x^2 + \cdots + (n-1)^2 x^{n-1} \right) \\ &&&= \frac{(n^2-n)x^{n+1} - 2(n^2-1)x^n + n(n+1)x^{n-1} -2}{(x-1)^3} - \frac{(n-1)x^n - nx^{n-1} +1}{(x-1)^2} \\[10pt] \implies &&1^2 &x + 2^2 x^2 + \cdots + (n-1)^2 x^{n-1} + n^2 x^n \\ &&&= x \left( \tfrac{(n^2-n)x^{n+1} - 2(n^2-1)x^n + n(n+1)x^{n-1} - 2 - ( (x-1)((x-1)x^n + nx^{n-1} + 1))}{(x-1)^3} \right)\\[10pt] \implies &&1^2 &x + 2^2 x^2 + \cdots + (n-1)^2 x^{n-1} + n^2 x^n \\ &&&= \frac{n^2 x^{n+3} - (2n^2 + 2n-1)x^{n+2} + (n+1)^2 x^{n+1} - x^2 - x}{(x-1)^3}. \end{align*}

    3. ( Note: This is a lot of algebra. I included as many steps as I thought appropriate, but that still has a lot going on in each step. If there is something that isn’t clear please leave a comment. Also, given all of these equations it seems likely I’ve made at least one typo / mathematical error. If you find one, please point it out.)