Be Sure Prior To shift me!

Algebra Level 3

1 2 × 2 3 × 3 4 × 4 5 × 5 6 × 6 7 × ⋯ = ? \dfrac{1}{{\color{#3D99F6} 2}}\times \dfrac{{\color{#3D99F6} 2}}{{\color{#D61F06} 3}}\times \dfrac{{\color{#D61F06} 3}}{{\color{#20A900} 4}}\times \dfrac{{\color{#20A900} 4}}{{\color{#E81990} 5}}\times \dfrac{{\color{#E81990} 5}}{{\color{#624F41} 6}}\times \dfrac{{\color{#624F41} 6}}{{\color{silver} 7}}\times \cdots = \,? David was given the above infinite product as his class assignment by his teacher, Daniel. He found difficult to answer this problem. To make him clear about it , Daniel arranged the product as 1 1 × 2 2 × 3 3 × 4 4 × 5 5 × 6 6 × 7 7 × ⋯ = ? \dfrac{1}{1}\times \dfrac{{\color{#3D99F6}2}}{{\color{#3D99F6} 2}}\times \dfrac{{\color{#D61F06} 3}}{{\color{#D61F06} 3}}\times \dfrac{{\color{#20A900} 4}}{{\color{#20A900} 4}}\times \dfrac{{\color{#E81990} 5}}{{\color{#E81990} 5}}\times \dfrac{{\color{#624F41} 6}}{{\color{#624F41} 6}}\times \dfrac{{\color{silver} 7}}{{\color{silver} 7}}\times \cdots = \,? At once he answered 1 1 . Is the rearrangement by Daniel correct?

Inspiration

No Yes

Naren Bhandari by Naren Bhandari , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Jordan Cahn
Nov 13, 2018

We can rewrite the given product as 1 2 × 2 3 × 3 4 × ⋯ = ∏ i = 1 ∞ i i + 1 \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots = \prod_{i=1}^\infty \frac{i}{i+1} By definition, ∏ i = 1 ∞ i i + 1 = lim ⁡ n → ∞ ∏ i = 1 n i i + 1 = lim ⁡ n → ∞ 1 n = 0 \begin{aligned} \prod_{i=1}^\infty \frac{i}{i+1} &= \lim_{n\to\infty}\prod_{i=1}^n \frac{i}{i+1} \\ &= \lim_{n\to\infty} \frac{1}{n} \\ &=0 \end{aligned} So the answer 1 1 is incorrect .

The rearrangement must also, therefore, be incorrect (since the product would, indeed, be 1 1 after rearranging). In general, you can't rearrange the denominators of an infinite product of fractions.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, it is but the theme of the problem about the arrangement of the terms in product above.

Naren Bhandari - 2 years, 6 months ago

Log in to reply

That was edited after I wrote my solution.

Jordan Cahn - 2 years, 6 months ago
Parth Sankhe
Nov 13, 2018

If you keep on crossing out the terms in the first series, you will eventually get 1 × 1 ∞ 1×\frac {1}{∞} , which would be 0.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Yes, it is but the theme of the problem about the arrangement of the terms in product above.

Naren Bhandari - 2 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...