Calculus

Common Misconceptions (Calculus)

Calculus Common Misconceptions: Level 2 Challenges

         

I'm going to prove that 1+2+22+23+⋯=−11 + 2 + 2^2 + 2^3 + \cdots = -1 . Is the following working correct?

Let AA denote the value of 1+2+22+23+⋯1 + 2 + 2^2 + 2^3 + \cdots , then

2A=2+22+23+24+⋯2A+1=1+2+22+23+24+⋯2A+1=AA=−1. \begin{aligned} 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{aligned}

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Pi Han Goh by Pi Han Goh

True or False?

lim⁡x→∞x+sin⁡xx+1=lim⁡x→∞ddx(x+sin⁡x)ddx(x+1)\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)}

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Brian Moehring by Brian Moehring

True or False?

1+2+3+4+5+6+⋯2+4+6+8+10+12+⋯=12 \dfrac{1 + 2 + 3 +4 + 5 + 6+ \cdots } {2 + 4 + 6 + 8 + 10 + 12 + \cdots } = \dfrac12

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Pi Han Goh 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?

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Justin Malme by Justin Malme

True or False?

Let S=11−12+13−14+15−16+⋯⇒12S=12−14+16−18+⋯ .\begin{aligned} S &= \color{#3D99F6}{\frac{1}{1}} \color{#D61F06}{-\frac{1}{2}} \color{#3D99F6}{+\frac{1}{3}} \color{#20A900}{-\frac{1}{4}} \color{#3D99F6}{+\frac{1}{5}} {\color{#D61F06}{- \frac{1}{6}}} \color{#333333}+ \cdots \\ \Rightarrow \frac{1}{2}S &= \color{#D61F06}{\frac{1}{2}} \color{#20A900}{- \frac{1}{4}} \color{#D61F06}{+ \frac{1}{6}} \color{#20A900}{- \frac{1}{8}} \color{#333333}+ \cdots. \end{aligned} Adding the two equations and regrouping gives 32S=(11+13+15+⋯)−2(14+18+112+⋯)=11−12+13−14+⋯=S⇒S=0.\begin{aligned} \frac{3}{2} S &= \color{#3D99F6}{\bigg ( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots \bigg )} - \color{#20A900}{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{aligned} Therefore, 11−12+13−14+⋯=0.\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = 0.

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Christopher Boo by Christopher Boo
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