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

by RoRi

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 the recursive sequence an+1 = (an + an-1) / 2

by RoRi

Let \{ a_n \} be the recursively defined sequence

    \[ a_{n+1} = \frac{a_n + a_{n-1}}{2} \qquad \text{for } n \geq 2, \]

where the first two terms are given a_1 and a_2.

  1. Assume that this sequence \{ a_n \} converges and compute its limit in terms of the initial terms a_1 and a_2.
  2. Prove that the sequences converges for every choice of a_1 and a_2. Assume a_1 < a_2.
  1. Assume the limit exists, say

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

    Then, we have

        \[ a_n + 2a_{n+1} = a_n + 2 \cdot \left( \frac{a_n + a_{n-1}}{2} \right) = a_{n-1} + 2a_n. \]

    Hence, the value a_n + 2a_{n+1} does not depend on n, and we have

        \[ a_n + 2a_{n+1} = a_1 + 2a_2. \]

    Now, taking the limit

        \begin{align*} && \lim_{n \to \infty} \left( a_n + 2a_{n+1} \right) &= \lim_{n \to \infty} (a_1 + 2a_2) \\[9pt] \implies && 3L &= a_1 + 2a_2 \\[9pt] \implies && L &= \frac{1}{3} a_1 + \frac{2}{3} a_2 \\[9pt] \implies && \lim_{n \to \infty} a_n &= \frac{1}{3} a_1 + \frac{2}{3} a_2. \end{align*}

  2. Proof. Assume a_1 < a_2 and let d = a_2 - a_1 > 0. Now, we claim for n \geq 2 that

        \[ a_{n+1} = a_n + (-1)^{n-1} \cdot \frac{a_2-a_1}{2^{n-1}}. \]

    We prove this claim by induction. For the case n = 2 we have

        \[ a_3 = \frac{a_2 + a_1}{2} = a_2 + \frac{a_1 - a_2}{2} = a_2 + (-1) \cdot \frac{a_2 - a_1}{2} = a_n + (-1)^{n-1} \cdot \frac{a_2 - a_1}{2^{n-1}}. \]

    So the statement is indeed true for the case n = 3. Now, assume the statement is true for some integer k \geq 2. Then we have

        \begin{align*} a_{k+2} &= \frac{a_{k+1} + a_k}{2} \\[9pt] &= \frac{ a_k + (-1)^{k-1} \cdot \frac{a_2-a_1}{2^{k-1}} + a_k}{2} &(\text{Ind. Hyp.}) \\[9pt] &= \frac{2^{k-1}a_k +(-1)^{k-1} (a_2 - a_1) + 2^{k-1} a_k}{2^k} \\[9pt] &= \frac{2^k a_k + (-1)^{k-1} (a_2 - a_1)}{2^k} \\[9pt] &= a_k + (-1)^{k-1} \cdot \frac{a_2-a_1}{2^k} \\[9pt] &= a_{k+1} - (-1)^{k-1} \frac{a_2-a_1}{2^{k-1}} + (-1)^{k-1} \cdot \frac{a_2-a_1}{2^k} &(\text{IH for term } a_k)\\[9pt] &= a_{k+1} - 2 \cdot (-1)^{k-1} \frac{a_2-a_1}{2^k} + (-1)^{k-1} \cdot \frac{a_2-a_1}{2^k} \\[9pt] &= a_{k+1} - (-1)^{k-1} \frac{a_2-a_1}{2^k} \\[9pt] &= a_{k+1} + (-1)^k \frac{a_2-a_1}{2^k}. \end{align*}

    Therefore, the proposed formula indeed holds for all integers n \geq 2. We then express this formula as a sum,

        \begin{align*} a_{n+1} &= a_n + (-1)^{n-1} \cdot \frac{a_2 - a_1}{2^{n-1}} \\[9pt] &= a_{n-1} + (-1)^{n-2} \cdot \frac{a_2 - a_1}{2^{n-2}} + (-1)^{n-1} \cdot \frac{a_2 - a_1}{2^{n-1}} \\[9pt] & \qquad \qquad \vdots \\[9pt] &= a_1 + \sum_{k=1}^n (-1)^{k-1} \cdot \frac{a_2 - a_1}{2^{n-1}}. \end{align*}

    So the terms of the sequence are the partial sums of this series. But, since a_2 - a_1 is a constant (for any given a_1 and a_2) this is a geometric series; hence, it converges. Therefore, the terms of the sequence \{ a_n \} converge to a finite limit. \qquad \blacksquare

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

by RoRi
  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

by RoRi
  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*}

Determine the convergence of the series 1 – n sin (1/n)

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} \left( 1 - n \sin \frac{1}{n} \right). \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges absolutely.

Proof. We know from the previous exercise (Section 10.20, Exercise #30) that the series

    \[ \sum_{n=1}^{\infty} \frac{\sin \frac{1}{n}}{n} \]

converges absolutely. Using the limit comparison test we have

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{1 - n \sin \frac{1}{n}}{\frac{1}{n} \sin \frac{1}{n}} \\[9pt] &= \lim_{n \to \infty} \frac{\frac{1}{n} - \sin \frac{1}{n}}{\frac{1}{n^2} \sin \frac{1}{n}}. \end{align*}

Now we consider these as functions of a real-variable and make the substitution n = \frac{1}{x}, and use L’Hopital’s rule three times to take the limit,

    \begin{align*} \lim_{n \to \infty} \frac{\frac{1}{n} - \sin \frac{1}{n}}{\frac{1}{n^2} \sin \frac{1}{n}} &= \lim_{x \to 0} \frac{x - \sin x}{x^2 \sin x} \\[9pt] &= \lim_{x \to 0} \frac{1 - \cos x}{2x \sin x + x^2 \cos x} \\[9pt] &= \lim_{x \to 0} \frac{sin x}{(2+x^2) \sin x + 4x \cos x} \\[9pt] &= \lim_{x \to 0} \frac{\cos x}{-2x \sin x + (6+x^2)\cos x} \\[9pt] &= \frac{1}{6}. \end{align*}

Therefore, since \sum b_n converges absolutely we have established the absolute convergence of

    \[ \sum_{n=1}^{\infty} \left( 1 - n \sin \frac{1}{n} \right). \qquad \blacksquare\]

Determine the convergence of the series 1 / n(1 + 1/2 + … + 1/n)

by RoRi

Consider the series

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

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series diverges.

Proof. We use the limit comparison test with the series \sum \frac{1}{n \log n}. Let

    \[ a_n = \frac{1}{n \log n}, \qquad b_n = \frac{1}{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)}. \]

Then, taking the limit we have

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)}{n \log n} \\[9pt] &= \lim_{n \to \infty} \frac{1 + \frac{1}{2} +\cdots + \frac{1}{n} }{\log n} \\[9pt] &= 1. \end{align*}

(The final equality follows from Example 3 on page 405 of Apostol, which tells us that this limit is 1.) Therefore, \sum a_n and \sum b_n either both converge or both diverge. But since

    \[ \sum_{n=1}^{\infty} \frac{1}{n \log n} \]

diverges we can then conclude that

    \[ \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)} \]

diverges as well. \qquad \blacksquare

Determine the convergence of the series (-1)n arctan (1 / (2n+1))

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n \arctan \frac{1}{2n+1}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges conditionally.

Proof. We can see that the series converges by the Leibniz rule since

    \[ \lim_{n \to \infty} \arctan \frac{1}{2n+1} = 0. \]

(Since \lim_{ x \to 0} \arctan x = 0). Furthermore, \arctan \frac{1}{2n+1} is monotonically decreasing since

    \[ \left( \arctan \frac{1}{2x+1} \right)' = \frac{1}{1 + \left( \frac{1}{2x+1} \right)^2} = \frac{-1}{1+2x + 2x^2} = \frac{-1}{(1+x)^2 + x^2} \]

is always negative.

Then, we can see that the convergence is conditional since

    \[ \sum_{n=1}^{\infty} \left| (-1)^n\arctan \left( \frac{1}{2n+1} \right) \right| = \sum_{n=1}^{\infty} \arctan \left( \frac{1}{2n+1} \right). \]

We use the limit comparison test with the series \sum \frac{1}{2n}. We have

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{\arctan \left(\frac{1}{2n+1} \right)}{\frac{1}{2n}} \\[9pt] &= \lim_{n \to \infty} \frac{ \frac{-1}{2n^2+2n+1}}{\frac{-1}{2n^2}} &(\text{L'Hopital's}) \\[9pt] &= \lim_{n \to \infty} \frac{2n^2}{2n^2+2n+1} \\[9pt] &= 1. \end{align*}

(Technically, to use L’Hopital’s and take this limit, I should look at the real-valued functions instead of the functions only taking values on the integers, and then say that this implies the limit of the integer-valued functions goes to 1.) By the limit comparison test we then know the two series either both converge or both diverge. Since \sum \frac{1}{2n} diverges we have

    \[ \sum_{n =1}^{\infty} \arctan \left( \frac{1}{2n+1} \right) \]

diverges as well. Therefore, the convergence of the series in the question is conditional. \qquad \blacksquare

Determine the convergence of the series (-1)n (n1/2 / (n+100))

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} (-1)^n\frac{\sqrt{n}}{n+100}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series is conditionally convergent.

Proof. By the Leibniz rule we know that if \{ a_n \} is a monotonic decreasing sequence with \lim_{n \to \infty} a_n = 0, then the alternating series \sum_{n=1}^{\infty} (-1)^{n-1} a_n converges. In this case we let

    \[ a_n = \frac{\sqrt{n}}{n+100}. \]

Then, \{ a_n \} is monotonic decreasing for n > 100. We can see this by considering

    \[ f(x) = \frac{\sqrt{x}}{x+100} \quad \implies \quad f'(x) = \frac{100-x}{2\sqrt{x} (100+x)^2}. \]

This is negative for x > 100; hence, f(x) is decreasing for x > 100. Since \{ a_n \} is the sequence of the values of f(x) on the integers we have \{ a_n \} is decreasing if n> 100. Further, we have

    \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt{n}}{100+n} = 0. \]

Thus, we write,

    \begin{align*} \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n+100} &= \sum_{n=1}^{100} (-1)^n \frac{\sqrt{n}}{n+100} + \sum_{n=101}^{\infty} (-1)^n \frac{\sqrt{n}}{n+100}. \end{align*}

This converges since the first term is some finite number (since it is a finite sum) and the second term converges by the Leibniz rule; hence, the sum of these two terms converges.

Finally, the convergence is conditional since

    \[ \sum_{n=1}^{\infty} \left| (-1)^n \frac{\sqrt{n}}{n+100} \right| = \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n+100}. \]

Letting b_n = \frac{1}{\sqrt{n}} we have

    \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{ \frac{\sqrt{n}}{n+100}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n}{n+100} = 1. \]

Hence, by the limit comparison test (and the fact that \sum b_n diverges) we conclude know that

    \[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n+100} \]

diverges. \qquad \blacksquare