Common Misconceptions (Calculus)
I'm going to prove that \(1 + 2 + 2^2 + 2^3 + \cdots = -1 \). Is the following working correct?
Let \(A \) denote the value of \(1 + 2 + 2^2 + 2^3 + \cdots \), then
\[ \begin{eqnarray} 2A &=& 2 + 2^2 + 2^3 + 2^4 + \cdots \\ 2A + 1 &=&1 + 2 + 2^2 + 2^3 + 2^4 + \cdots \\ 2A + 1 &=&A \\ A &=&-1. \end{eqnarray} \]
by
Pi Han Goh
True or False?
\[\large \lim_{x\rightarrow\infty} \frac{x + \sin x}{x+1} = \lim_{x\rightarrow\infty} \frac{\tfrac{d}{dx}(x + \sin x)}{\tfrac{d}{dx}(x+1)}\]
by
Brian Moehring
True or False?
\[ \dfrac{1 + 2 + 3 +4 + 5 + 6+ \cdots } {2 + 4 + 6 + 8 + 10 + 12 + \cdots } = \dfrac12 \]
by
Pi Han Goh
Suppose you are standing one meter from a wall, and you begin walking towards the wall. With each step, you cover half the distance between you and the wall. If you do this at a rate of one meter per second, will you ever reach the wall?
by
Justin Malme
True or False?
Let \[\begin{align} S &= \color{Blue}{\frac{1}{1}} \color{Red}{-\frac{1}{2}} \color{Blue}{+\frac{1}{3}} \color{Green}{-\frac{1}{4}} \color{Blue}{+\frac{1}{5}} {\color{Red}{- \frac{1}{6}}} \color{black}+ \cdots \\ \Rightarrow \frac{1}{2}S &= \color{Red}{\frac{1}{2}} \color{Green}{- \frac{1}{4}} \color{Red}{+ \frac{1}{6}} \color{Green}{- \frac{1}{8}} \color{black}+ \cdots. \end{align}\] Adding the two equations and regrouping gives \[\begin{align} \frac{3}{2} S &= \color{Blue}{\bigg ( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots \bigg )} - \color{Green}{2 \bigg ( \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \cdots \bigg )} \\ &= \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = S \\\\ \Rightarrow S &= 0. \end{align}\] Therefore, \[\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = 0.\]
by
Christopher Boo