Finite or Infinite?

Level 1

Is ( 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 3 \frac{1}{3} + 1 4 \frac{1}{4} + 1 5 \frac{1}{5} + 1 6 \frac{1}{6} + 1 7 \frac{1}{7} + 1 8 \frac{1}{8} + ... ) finite or infinite? Prove why it is either finite or infinite.

Finite Infinite

Joshua Lowrance by Joshua Lowrance , 18, USA

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2 solutions

X X
Aug 24, 2018

Although we know th harmonic series 1 + 1 2 + 1 3 + . . . + 1 n ≈ ln ⁡ n + γ 1+\frac12+\frac13+...+\frac1n\approx\ln n+\gamma , I will still write a prove.

1 1 = 1 1 \dfrac11=\dfrac11

1 2 + 1 3 + 1 4 = 1 4 + 1 4 + 1 3 + 1 4 > 1 \dfrac12+\dfrac13+\dfrac14=\dfrac14+\dfrac14+\dfrac13+\dfrac14>1

1 5 + 1 6 + . . . + 1 25 = 1 25 + 1 25 + 1 25 + 1 25 + 1 25 + 1 6 + 1 7 + . . . + 1 25 > 1 \dfrac15+\dfrac16+...+\dfrac1{25}=\dfrac1{25}+\dfrac1{25}+\dfrac1{25}+\dfrac1{25}+\dfrac1{25}+\dfrac16+\dfrac17+...+\dfrac1{25}>1

⋮ \vdots

1 n + 1 n + 1 + . . . + 1 n 2 > 1 \dfrac1n+\dfrac1{n+1}+...+\dfrac1{n^2}>1

So, 1 + 1 2 + 1 3 + 1 4 + . . . > 1 + 1 + 1 + 1 + . . . = ∞ 1+\dfrac12+\dfrac13+\dfrac14+...>1+1+1+1+...=\infty

3 Helpful 0 Interesting 0 Brilliant 0 Confused

interesting proof, and great use of LaTeX \LaTeX{}

chase marangu - 2 years, 9 months ago
Joshua Lowrance
Aug 23, 2018

( 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 3 \frac{1}{3} + 1 4 \frac{1}{4} + 1 5 \frac{1}{5} + 1 6 \frac{1}{6} + 1 7 \frac{1}{7} + 1 8 \frac{1}{8} + ... ) > > ( 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 4 \frac{1}{4} + 1 4 \frac{1}{4} + 1 8 \frac{1}{8} + 1 8 \frac{1}{8} + 1 8 \frac{1}{8} + 1 8 \frac{1}{8} + ... ) = ( 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 2 \frac{1}{2} + 1 2 \frac{1}{2} + 1 2 \frac{1}{2} + 1 2 \frac{1}{2} + 1 2 \frac{1}{2} + 1 2 \frac{1}{2} + ... ) = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + . . . 1+1+1+1+1+1+1+1+... = ∞ \infty

So therefore, ( 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 3 \frac{1}{3} + 1 4 \frac{1}{4} + 1 5 \frac{1}{5} + 1 6 \frac{1}{6} + 1 7 \frac{1}{7} + 1 8 \frac{1}{8} + ... ) = ∞ \infty

2 Helpful 0 Interesting 0 Brilliant 0 Confused

So your saying (thing) ≥ 1+(thing) therefore (thing) ≥ 1+1+1+⋯ therefore thing ≥ ∞ but we know (thing) ≯ ∞ so (thing) = ∞

chase marangu - 2 years, 9 months ago

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Yes that is correct

Joshua Lowrance - 2 years, 9 months ago

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interesting. I like to call phi=√1+√1+√1+⋯ etc. "infinitely iterated recursive substitution" has a ring to it. There is some practical applications of it but most people find it as just a fun little thing for recreational math. There really are practical applications of it tho this proof is one of them

chase marangu - 2 years, 9 months ago

Ellipsis doesn't necessarily mean infinite. If there is a formula contained that writes a recurring law instead or with the ellipsis then this may be infinite.

R S C - 1 year, 1 month ago

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@R S C bro, I literally just said that. Also, You just posted on a litteraly 2-year-old comment. Also, it was implied in my syntax that the ellipsis meant "Put X Here Recursively" like, get with it bro

chase marangu - 1 year, 1 month ago

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@chase marangu thing inside parenthesis isn't exactly clear "bro"

R S C - 1 year, 1 month ago

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