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Determine the convergence or divergence of f(n) = (n+1)1/2 – n1/2

Consider the function f(n) defined by

    \[ f(n) = \sqrt{n+1} - \sqrt{n}. \]

Determine whether the sequence \{ f(n) \} converges or diverges, and if it converges find the limit.


First, we use multiply the numerator and denominator by \sqrt{n+1} + \sqrt{n},

    \begin{align*}  f(n) &= \sqrt{n+1} - \sqrt{n} \\[9pt]  &= \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} \\[9pt]  &= \frac{n+1 - n}{\sqrt{n+1} + \sqrt{n}} \\[9pt]  &= \frac{1}{\sqrt{n+1} + \sqrt{n}}. \end{align*}

Then, since \sqrt{n+1} > \sqrt{n} we have

    \[ 0 < \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2 \sqrt{n}}. \]

Since we know from property (10.9) on page 380 of Apostol that

    \[ \lim_{n \to \infty} \frac{1}{n^{\alpha}} = 0 \qquad \text{if } \alpha > 0 \]

we then have

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

Hence, by the squeeze theorem we have

    \[ \lim_{n \to \infty} f(n) = 0. \]

Therefore the sequence \{ f(n) \} converges with limit 0.

In two different ways, prove the limit as x tends to 0 of (log (1+x))/x is 1

The following limit equation is valid:

    \[ \lim_{x \to 0} \frac{\log (1+x)}{x} = 1. \]

Prove this in the following two ways:

  1. Using the definition of the derivative L'(1);
  2. Using the previous exercise (Section 6.9, Exercise #28).

  1. Proof. First, we know from Theorem 6.1 part (b) (page 229 of Apostol) that

        \[ L'(x) = \frac{1}{x} \qquad \text{for all } x > 0. \]

    Therefore, L'(1) = 1. Then, we recall the definition of the derivative of a function

        \[ f'(x) = \lim_{x \to 0} \frac{f(x+h) - f(x)}{h} . \]

    So, using the definition of the derivative and the fact that for x > 0 we have L(x) = \log x,

        \begin{align*}  L'(x) &= \lim_{h \to 0} \frac{L(x+h) - L(x)}{h} \\   &= \lim_{h \to 0} \frac{\log(x+h) - \log(x)}{h}. \end{align*}

    Evaluating at 1 we have

        \begin{align*}  1 = L'(1) &= \lim_{h \to 0} \frac{\log (1+h) - \log 1}{h} \\  &= \lim_{h \to 0} \frac{\log (1+h)}{h} \\  &= \lim_{x \to 0} \frac{\log (1+x)}{x} \end{align*}

    since the variable name is unimportant. Thus, we have established the requested formula,

        \[ \lim_{x \to 0} \frac{\log (1+x)}{x} = 1. \qquad \blacksquare \]

  2. Proof. From the previous exercise (Section 6.9, Exercise #28) we know

        \[ 1 - \frac{1}{x} < \log x < x - 1 \qquad \text{for all } x > 0 \text{ with } x \neq 1. \]

    If x > 0 then x+1 > 0 as well, so the inequalities still hold with x + 1,

        \[ 1 - \frac{1}{x+1} < \log (x+1) < x. \]

    Multiplying all of the terms by \frac{1}{x} (since x > 0 we may do this without reversing the inequalities),

        \[ \frac{1}{x} - \frac{1}{x(x+1)} < \frac{\log(x+1)}{x} < 1. \]

    Since \frac{1}{x+1} = \frac{1}{x} - \frac{1}{x(x+1)} we then have,

        \[ \frac{1}{x+1} < \frac{\log(x+1)}{x} < 1. \]

    Finally, since

        \[ \lim_{x \to 0} \frac{1}{1+x} = 1 , \qquad \text{and} \qquad \lim_{x \to 0} 1 = 1, \]

    we apply the squeeze theorem (Theorem 3.3, page 133 of Apostol) to conclude,

        \[ \lim_{x \to 0} \frac{\log(1+x)}{x} = 1. \qquad \blacksquare\]