Almost all numbers present

Calculus Level 4

1 1 × 2 × 3 + 1 5 × 6 × 7 + 1 9 × 10 × 11 + … = ? \frac {1}{1 \times 2 \times 3} + \frac 1 {5 \times 6 \times 7} + \frac 1 {9 \times 10 \times 11} + \ldots = \ ?

Details and Assumptions :

  • You are encouraged to find the exact form of the infinite sum above.


The answer is 0.173.

Keshav Tiwari by Keshav Tiwari , 23, 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.

3 solutions

Tijmen Veltman
Apr 8, 2015

One can easily check that 1 ( x − 1 ) × x × ( x + 1 ) = 1 2 x − 1 − 1 x + 1 2 x + 1 \frac1{(x-1)\times x\times (x+1)}=\frac{\frac12}{x-1}-\frac1x+\frac{\frac12}{x+1} . This gives:

1 1 × 2 × 3 + 1 5 × 6 × 7 + … \frac1{1\times 2\times 3}+\frac1{5\times 6\times 7}+\dots

= 1 2 ( 1 1 − 1 2 − 1 2 + 1 3 + 1 5 − 1 6 − 1 6 + 1 7 + … ) = \frac12\left(\frac11-\frac12-\frac12+\frac13+\frac15-\frac16-\frac16+\frac17+\dots\right)

= 1 2 ( 1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 + … ) + 1 2 ( − 1 2 + 1 4 − 1 6 + 1 8 − … ) = \frac12\left(\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\right) +\frac12\left( -\frac12+\frac14-\frac16+\frac18-\dots\right)

= 1 2 ( 1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 + … ) − 1 4 ( 1 1 − 1 2 + 1 3 − 1 4 + … ) = \frac12\left(\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\right) -\frac14\left( \frac11-\frac12+\frac13-\frac14+\dots\right)

= 1 4 ( 1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 + … ) = \frac14\left(\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots\right)

= 1 4 log ⁡ 2 ≈ 0.173 =\frac14\log 2 \approx \boxed{0.173} .

For more information on the final sum, see here .

12 Helpful 0 Interesting 0 Brilliant 0 Confused

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Abhimanyu Singh
May 25, 2015

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...