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Prove that the recursive sequence 1 / xn+2 = 1 / xn+1 + 1 / xn converges

Prove that the sequence \{ x_n \} be defined by the recursive relationship,

    \[ x_0 = 1, \qquad x_1 = 1, \qquad \frac{1}{x_{n+2}} = \frac{1}{x_{n+1}} + \frac{1}{x_n} \]

converges and find the limit of the sequence.


Proof. First, we show that the sequence \{ x_n \} is monotonically decreasing for all n \geq 1. For the base case we have x_1 = 1 and

    \[ \frac{1}{x_2} = \frac{1}{1} + \frac{1}{1} = 2 \quad \implies \quad x_2 = \frac{1}{2}. \]

Hence, x_2 < x_1. Assume then that for all positive integers up to some k we have x_{k+1} < x_k. Then,

    \begin{align*}  && \frac{1}{x_{k+2}} &= \frac{1}{x_{k+1}} + \frac{1}{x_k} \\[9pt]  \implies && \frac{1}{x_{k+2}} &> \frac{1}{x_{k+1}} + \frac{1}{x_{k+1}} &(x_{k+1} < x_k \implies \frac{1}{x_{k+1}} > \frac{1}{x_k}) \\[9pt]  \implies && \frac{1}{x_{k+2}} &> \frac{2}{x_{k+1}} \\[9pt]  \implies && x_{k+2} &< \frac{x_{k+1}}{2} \\[9pt]  \implies && x_{k+2} &< x_{k+1}. \end{align*}

Thus, the sequences is monotonically decreasing. The sequence is certainly bounded below since all of the terms are greater than 0. Therefore, the sequence converges. \qquad \blacksquare

To compute the limit of the sequence, assume the sequence converges to a finite limit L (justified since we just proved that it does indeed converge). Therefore,

    \begin{align*}  && \lim_{n \to \infty} x_n &= L \\[9pt]  \implies && \lim_{n \to \infty} \frac{1}{x_n} &= \frac{1}{L} \\[9pt]  \implies && \lim_{n \to \infty} \left( \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} \right) &= \frac{1}{L} \\[9pt]  \implies && \frac{1}{L} + \frac{1}{L} &= \frac{1}{L} \\[9pt]  \implies && 2L &= L \\[9pt]  \implies && L &= 0. \end{align*}

Prove that the recursive sequence xn+1 = (1+xn)1/2 converges

Prove that the sequence \{ x_n \} whose terms are defined recursively by

    \[ x_1 = 1, \qquad x_{n+1} = \sqrt{1+x_n} \]

converges, and compute the limit of the sequence.


Proof. To show the sequence converges we show that it is monotonically increasing and bounded above. To see that it is monotonically increasing we use induction to prove that

    \[ x_{n+1} > x_n \qquad \text{for all } n \in \mathbb{Z}_{>0}. \]

For the case n = 1 we have

    \[ x_{n+1} = x_2 = \sqrt{2}, \qquad \text{and} \qquad x_n = x_1 = 1. \]

Since \sqrt{2}  >1, the statement holds in the case n =1. Assume then that the statement holds for some positive integer k. Then,

    \begin{align*}  x_{k+2} - x_{k+1} &= \sqrt{1+x_{k+1}} - \sqrt{1+x_k}\\[9pt]  &> 0 \end{align*}

since x_{k+1} > x_k by the induction hypothesis. Hence, x_{k+2} > x_{k+1} so by induction x_{n+1} > x_n for all positive integers n. Hence, the sequence is monotonically increasing.
Next we use induction again to prove the sequence is bounded above by 2. For n = 1 we have x_1 = 1 < 2 so the hypothesis holds. Assume then that x_k < 2 for all positive integers up to k. Then,

    \begin{align*}  x_{k+1} &= \sqrt{1+x_k} \\  &\leq \sqrt{1+2} \\  &\leq 2. \end{align*}

Hence, x_n < 2 for all positive integers n.
This shows that the sequence converges.\qquad \blacksquare

To compute the limit, assume the sequence converges to a number L (we just proved that it converges, so this assumption is valid). Then we have

    \begin{align*}  \lim_{n \to \infty} x_{n+1} &= L & \implies && \lim_{n \to \infty} \sqrt{1+x_n} &= L \\[9pt]  && \implies && \sqrt{1+L} &= L \\[9pt]  && \implies && L^2 - L - 1 &= 0 \\[9pt]  && \implies && L &= \frac{1 + \sqrt{5}}{2}. \end{align*}

(We can discard the negative solution since to the quadratic at the end since the sequence is certainly all positive terms.)

Compute the limit of (1+xn)1/n and (an + bn1/n

  1. Consider the limit

        \[ \lim_{n \to \infty} (1+x^n)^{\frac{1}{n}}. \]

    For 0 < x < 1 prove that this limit exists and compute the limit.

  2. For positive real numbers a and b, Consider the limit

        \[ \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}}. \]

    Compute this limit.


  1. Proof. We use the squeeze theorem to prove existence and compute the value of the limit. Since 0<x<1 we have

        \[ (1+0^n)^{\frac{1}{n}} < (1+x^n)^{\frac{1}{n}} < (1+1^n)^{\frac{1}{n}} \]

    for all positive integers n. Then we have

        \begin{align*}  \lim_{n \to \infty} (1+0^n)^{\frac{1}{n}} &= \lim_{n \to \infty} 1^{\frac{1}{n}} = 1 \\[9pt]  \lim_{n \to \infty} (1+1^n)^{\frac{1}{n}} &= \lim_{n \to \infty} 2^{\frac{1}{n}} = 1. \end{align*}

    (We know the second limit from this previous exercise (Section 10.4, Exercise #9).) Therefore, by the squeeze theorem we have

        \[ \lim_{n \to \infty} \left( 1 + x^n \right)^{\frac{1}{n}} = 1. \qquad \blacksquare \]

  2. If a > b then we have 0 < \frac{a}{b} < 1 and so, using part (a),

        \begin{align*}  \lim_{n \to \infty} \left( a^n + b^n \right)^{\frac{1}{n}} &= \lim_{n \to \infty} a \cdot \left( 1 + \left( \frac{b}{a} \right)^n \right)^{\frac{1}{n}} \\[9pt]  &= a. \end{align*}

    On the other hand if b > a then we have 0 < \frac{a}{b} < 1 and so by part (a) again,

        \begin{align*}  \lim_{n \to \infty} \left( a^n + b^n \right)^{\frac{1}{n}} &= \lim_{n \to \infty} b \cdot \left( 1 + \left(\frac{a}{b}\right)^n \right)^{\frac{1}{n}} \\[9pt]  &= b. \end{align*}

    If a = b then

        \begin{align*}  \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}} &= a \cdot \lim_{n \to \infty} \left( 1 + \left( \frac{b}{a} \right)^n \right)^{\frac{1}{n}} \\[9pt]  &= a \cdot \lim_{n \to \infty} 2^{\frac{1}{n}} \\[9pt]  &= a = b. \end{align*}

Compute limits of (n+1)c – nc for real values of c

  1. Compute the limit

        \[ \lim_{n \to \infty} a_n \qquad \text{where} \qquad a_n = \sqrt{n+1} - \sqrt{n}. \]

  2. Compute the limit

        \[ \lim_{n \to \infty} a_n \qquad \text{where} \qquad a_n = (n+1)^c - n^c \]

    for c \in \mathbb{R}. Determine the values for which the sequences diverges and for which it converges and compute the values of the limits in the convergent case.


  1. We multiply and divide the terms by \sqrt{n+1} + \sqrt{n} to get,

        \begin{align*}  \lim_{n \to \infty} \left( \sqrt{n+1} - \sqrt{n} \right) &= \lim_{n \to \infty} \left( \frac{\left( \sqrt{n+1} - \sqrt{n} \right)\left( \sqrt{n+1} + \sqrt{n} \right)}{\sqrt{n+1} + \sqrt{n}}\right) \\[9pt]  &= \lim_{n \to \infty} \frac{1}{\sqrt{n+1} + \sqrt{n}} \\[9pt]  &= 0. \end{align*}

  2. Observe that

        \[ (n+1)^c - n^c = c \int_n^{n+1} \frac{1}{x^{1-c}} \, dx \]

    since

        \[ c \int_n^{n+1} \frac{1}{x^{1-c}} \,dx = c \cdot \left( \frac{x^c}{c} \Bigr \rvert_n^{n+1} \right) = (n+1)^c - n^c. \]

    But then, if 0 < c < 1, the function inside the integral is a decreasing function; hence,

        \[ c \int_n^{n+1} \frac{1}{x^{1-c}} \, dx \leq \frac{c}{n^{1-c}}. \]

    Since this limit goes to 0 for 0 < c < 1 (since 1-c > 0) we have that the sequence converges to 0 for these values of c.
    If c > 1 then the integrand is increasing, and so,

        \[ c \int_n^{n+1} \frac{1}{x^{1-c}} \, dx \geq \frac{c}{n^{1-c}}. \]

    In this case the limit of \frac{c}{n^{1-c}} as n \to \infty diverges to +\infty (since 1-c < 0). Hence, the sequence does not converge.
    Finally, if c= 1 then

        \[ \lim_{n \to \infty} ((n+1)^c - n^c) = \lim_{n \to \infty} 1 = 1. \]

    Putting this all together we have

        \begin{align*}  \{ a_n \} &\text{ converges to } 0 \text{ for } 0 < c < 1 \\[9pt]  \{ a_n \} & \text{ converges to } 1 \text{ for } c = 1 \\[9pt]  \{ a_n \} & \text{ diverges for } c > 1.  \end{align*}

Establish the given limit relations

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*}

Prove some statements about integrals of bounded monotonic increasing functions

Consider a bounded, monotonic, real-valued function f on the interval [0,1]. The define sequences

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

  1. Prove that

        \[ s_n \leq \int_0^1 f(x) \, dx \leq t_n \]

    and that

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

  2. Prove that the two sequences \{ s_n \} and \{ t_n \} converge to \int_0^1 f(x) \, dx.
  3. State and prove a generalization of the above to interval [a,b].

  1. Proof.First, we define two step functions,

        \[ s(x) = f \left( \frac{[nx]}{n} \right), \qquad t(x) = f \left( \frac{[nx+1]}{n} \right) \]

    where [x] denotes the greatest integer less than or equal to x. Then we define a partition of [0,1],

        \[ P = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \ldots, 1 \right\}. \]

    For any x_{k_1} \leq x < x_k we have

        \[ s(x) = f \left( \frac{k-1}{n} \right), \qquad t(x) = f \left( \frac{k}{n} \right). \]

    So, s(x) and t(x) are constant on the open subintervals of the partition P.
    Since f(x) is monotonically increasing and s(x) \leq f(x) \leq t(x) for all x (by the definition of s and t) we have

        \begin{align*}  &&\int_0^1 s(x) \, dx \leq \int_0^1 f(x) \, dx \leq \int_0^1 t(x) \, dx \\[9pt]  \implies && \sum_{k=0}^{n-1} s_k (x_k - x_{k-1} \leq \int_0^1 f(x) \, dx \leq \sum_{k=0}^{n-1} t_k (x_k - x_{k-1}) && (\text{Def step function integral}) \\[9pt]  \implies && \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k+1}{n} \right) &&(x_k - x_{k-1} = \frac{1}{n} \forall k) \\[9pt]  \implies && \fract{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=1}^nf \left( \frac{k}{n} \right) \\[9pt]  \implies && s_n \leq \int_0^1 f(x) \, dx \leq t_n && \text{for all } n \\[9pt]  \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq t_n - s_n \\[9pt]  \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n}. \qquad \blacksquare \end{align*}

  2. Proof. From part (a) we have

        \[ 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n} \quad \implies \quad \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - s_n \right)= 0 \]

    since \lim_{n \to \infty} \frac{f(1)- f(0)}{n} = 0. Since \int_0^1 f(x) \, dx does not depend on n we have

        \[ \int_0^1 f(x) \, dx  - \lim_{n \to \infty} s_n = 0 \quad \implies \quad \lim_{n \to \infty} s_n = \int_0^1 f(x) \, dx. \]

    Therefore,

        \begin{align*}   && s_n \leq \int_0^1 f(x) \, dx \leq t_n \\[9pt]  \implies && s_n - t_n \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt]  \implies && \frac{f(0)-f(1)}{n} \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt]  \implies && \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - t_n \right) = 0 \\[9pt]  \implies && \lim_{n \to \infty} t_n = \int_0^1 f(x) \, dx. \qquad \blacksquare \end{align*}

  3. Claim: If f is a real-valued function that is monotonic increasing and bounded on the interval [a,b], then

        \[ \lim_{n \to \infty} s_n = \int_a^b f(x) \, dx, \qquad \text{and} \qquad \lim_{n \to \infty} t_n = \int_a^b f(x) \, dx \]

    for s_n and t_n defined as follows:

        \[ s_n = \frac{b-a}{n} \cdot sum_{k=0}^{n-1} f \left( a + k \frac{b-a}{n} \right), \qquad t_n = \frac{b-a}{n} \cdot \sum_{k=1}^n f \left( a + k \frac{b-a}{n} \right). \]

    Proof. Let

        \[ P = \left\{ a, a + \frac{b-a}{n}, a + \frac{2(b-a)}{n}, \ldots, a + \frac{n(b-a)}{n} = b \right\}  \]

    be a partition of the interval [a,b]. Then, define step functions s(x) and t(x) with s(x) = f(x_{k-1}) and t(x) = f(x_k) for x_{k-1} \leq x < x_k. By these definitions we have s(x) \leq f(x) \leq t(x) for all x \in [a,b] (since f is monotonic increasing). Since f is bounded and monotonic increasing it is integrable, and

        \begin{align*}  && \int_a^b s(x) \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b t(x) \, dx \\[9pt]  \implies && \sum_{k=1}^n s_k (x_k - x_{k-1}) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k (x_k - x_{k-1}) \\[9pt]  \implies && \sum_{k=1}^n s_k \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k \left( \frac{b-a}{n} \right). \end{align*}

    And, since s_k = f(x_{k-1} = f \left( a + \frac{(k-1)(b-a)}{n} \right), and t_k = f(x_k) = f \left( a + \frac{k(b-a)}{n} \right) we have

        \begin{align*}  \implies && \sum_{k=1}^n f \left( a + \frac{(k-1)(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n f \left( a + \frac{k(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \\[9pt]  \implies && \left( \frac{b-a}{n} \right) \sum_{k=0}^{n-1} f \left( a+ \frac{k(b-a)}{n} \right) \leq \int_a^b f(x) \, dx \leq \left( \frac{b-a}{n} \right) \sum_{k=1}^n f \left( a+ \frac{k(b-a)}{n} \right). \qquad \blacksquare \end{align*}

Prove that the limit of a product is the product of the limits for sequences with finite limits

Let

    \[ \lim_{n \to \infty} a_n = A. \]

From the previous two exercises here and here (Section 10.4, Exercises #30 and #31) we know that

    \[ \lim_{n \to \infty} a_n = 0 \quad \implies \quad \lim_{n \to \infty} a_n^2 = 0 \]

and that

    \[ \lim_{n \to \infty} a_n = A, \ \ \lim_{n \to \infty} b_n = B \quad \implies \quad \lim_{n \to \infty} (a_n + b_n) = A+B. \]

Use these results to prove that

    \[ \lim_{n \to \infty} a_n^2 = A^2. \]

Then use the identity

    \[ 2 a_n b_n = (a_n + b_n)^2 - a_n^2 - b_n^2 \]

to prove that

    \[ \lim_{n \to \infty} (a_n b_n) = AB. \]


Proof. First,

    \begin{align*}  \lim_{n \to \infty} a_n = A && \implies && \lim_{n \to \infty} (a_n - A) &= 0 \\  && \implies && \lim_{n \to \infty} (a_n - A)^2 &= 0 (\text{Ex. } 30)\\  && \implies && \lim_{n \to \infty} (a_n^2 - 2Aa_n + A^2) &= 0 \\  && \implies && \lim_{n \to \infty} a_n^2 - 2A \lim_{n \to \infty} a_n + A^2 &= 0 \\  && \implies && \lim_{n \to \infty} a_n^2 &= A^2. \qquad \blacksquare \end{align*}

Proof. Letting

    \[ \lim_{n \to \infty} a_n = A \qquad \text{and} \qquad \lim_{n \to \infty} b_n = B \]

we have from Exercise #31,

    \[ \lim_{n \to \infty} (a_n + b_n) = A+ B. \]

Then by Exercise #30,

    \[ \lim_{n \to \infty} (a_n + b_n)^2 = (A+B)^2 = A^2 + 2AB + B^2. \]

Using the identity 2a_n b_n = (a_n + b_n)^2 - a_n^2 - b_n^2 we then have

    \begin{align*}  &&\lim_{n \to \infty} (a_n + b_n)^2 &= A^2 + 2AB + B^2 \\  \implies && \lim_{n \to \infty} (2a_n b_n + a_n^2 + b_n^2) &= A^2 + 2AB + B^2 \\  \implies && 2 \lim_{n \to \infty} a_n b_n + \lim_{n \to \infty} a_n^2 + \lim_{n \to \infty} b_n^2 &= A^2 + 2AB + B^2 \\  \implies && 2 \lim_{n \to \infty} a_n b_n &= 2AB \\  \implies && \lim_{n \to \infty} a_n b_n &= AB. \qquad \blacksquare \end{align*}

Prove that sequences with finite limits are linear with respect to taking limits

Let

    \[ \lim_{n \to \infty} a_n = A \qquad \text{and} \qquad \lim_{n \to \infty} b_n = B. \]

Prove directly from the definition of the limit that

    \[ \lim_{n \to \infty} (a_n + b_n) = A + B \qquad \text{and} \qquad \lim_{n \to \infty} (ca_n) = cA \]

for any constant c.


Proof. Let \varepsilon >0 be given. Since \lim_{n \to \infty} a_n = A and \lim_{n \to \infty} b_n = B we know there exist positive integers N_a and N_b such that

    \[ |a_n - A| < \frac{\varepsilon}{2} \qquad \text{and} \qquad |b_m - B| < \frac{\varepsilon}{2} \]

for all n \geq N_a and all m \geq N_b. Let N = \max \{ N_a, N_b \}. Then, for all n \geq N we have

    \begin{align*}  &&|a_n - A| + |b_n - B| &< \varepsilon \\  \implies && |a_n + b_n - (A+B)| &< \varepsilon. \end{align*}

Hence,

    \[ \lim_{n \to \infty} (a_n + b_n) = A+B. \qquad \blacksquare \]

Proof. Let \varepsilon > 0 be given. Since \lim_{n \to \infty} a_n = A we know there exists a positive integer N such that

    \[ |a_n - A| < \frac{\varepsilon}{|c|} \qquad \text{for all } n \geq N. \]

Thus,

    \begin{align*}  &&|c| \cdot |a_n - A| &< |c| \cdot \frac{\varepsilon}{|c|} \\[9pt]  \implies && | ca_n - cA| &< \varepsilon \\[9pt]  \implies && \lim_{n \to \infyt} ca_n &= cA. \qquad \blacksquare \end{align*}