Exercise 6.4.6

Question

Let
  $$
  f(x)=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+4}-\cdots .
  $$
  Show $f$ is defined for all $x>0$. Is $f$ continuous on $(0, \infty)$ ? How about differentiable?

Ulisse Mini · Accepted Answer

$f(x)$ converges for any $x > 0$ by the Alternating Series Test. Since $f$ converges we are free to use associativity to group the terms in pairs, from which we get
 $$ \begin{aligned}
    f(x) &= \left(\frac{1}{x}-\frac{1}{x+1}\right) + \left(\frac{1}{x+2}-\frac{1}{x+3}\right)+\cdots \
    &= \frac{1}{x^2 + x} + \frac{1}{(x+2)^2 + (x+2)}+ \cdots \
    &< \frac{1}{x^2} + \frac{1}{(x+2)^2} + \cdots
 \end{aligned}
    $$

Exercise 6.4.6

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Exercise 6.4.6

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