Capitalize the loss

Calculus Level 1

( 1 2 2 + 1 3 2 + 1 4 2 + … ) + ( 1 2 3 + 1 3 3 + 1 4 3 + … ) + ( 1 2 4 + 1 3 4 + 1 4 4 + … ) + ⋯ \left(\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\ldots\right) + \\ \hspace{1cm}\left(\dfrac1{2^3}+\dfrac1{3^3}+\dfrac1{4^3} + \ldots\right) + \\ \hspace{2.5cm}\left(\dfrac1{2^4}+\dfrac1{3^4}+\dfrac1{4^4}+\ldots\right)+\cdots

What is the value of the series above?


The answer is 1.

View wiki
Danish Ahmed by Danish Ahmed , 24, 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.

10 solutions

Danish Ahmed
Apr 21, 2015

Regroup the expression as shown:

( 1 2 2 + 1 2 3 + 1 2 4 + ⋯ ) + ( 1 3 2 + 1 3 3 + 1 3 4 + ⋯ ) + ( 1 4 2 + 1 4 3 + 1 4 4 + ⋯ ) + ⋯ \left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots\right)+\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\cdots\right)+\left(\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots\right)+\cdots .

Summing each of the geometric series,

a + a r + a r 2 + a r 3 + ⋯ = a 1 − r a+ar+ar^2+ar^3+\cdots=\dfrac{a}{1-r} Provided − 1 < r < 1 -1<r<1

we get

1 2 + 1 6 + 1 12 + ⋯ = ( 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ = 1 \dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\cdots=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\cdots=\boxed{1} .

101 Helpful 6 Interesting 11 Brilliant 9 Confused

Moderator note:

Great, what can we conclude for the value ∑ k = 2 ∞ ( ζ ( k ) − 1 ) \displaystyle \sum_{k=2}^\infty \left ( \zeta(k) - 1 \right ) ?

Each (infinite) group of terms is also a special value of the zeta function shifted by 1 , 1, e.g. 1 2 3 + 1 3 3 + 1 4 3 + . . . = ζ ( 3 ) − 1. \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + ... = \zeta(3) - 1. So the series can be written as ∑ k = 2 ∞ [ ζ ( k ) − 1 ] \sum_{k=2}^\infty [\zeta(k)-1] And since Danish proved the value to be 1 , 1, it is trivial to say ∑ k = 2 ∞ [ ζ ( k ) − 1 ] = 1 \sum_{k=2}^\infty [\zeta(k)-1] = 1

Dreama Williams - 6 years, 1 month ago

I got the first part right, but when I got to the second series:(1/2+1/6+1/12+1/20....) I didn't know what to do anymore because it was neither ap or gp and after reading this solution I still don't know how exactly did you handle that part?

simon maxwell - 4 years, 10 months ago

Log in to reply

Are you familiar with partial fractions and telescoping series and some basic calculus? If yes, here's a formal way to evaluate that series: 1 2 + 1 6 + 1 12 + 1 20 + … = lim ⁡ n → ∞ ∑ k = 1 n 1 k ( k + 1 ) = lim ⁡ n → ∞ ∑ k = 1 n ( 1 k − 1 k + 1 ) = lim ⁡ n → ∞ ( 1 − 1 n + 1 ) = 1 \frac 12+\frac 16+\frac 1{12}+\frac 1{20}+\ldots=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)}=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac 1k-\frac 1{k+1}\right)=\lim_{n\to\infty}\left(1-\frac 1{n+1}\right)=1

Prasun Biswas - 4 years, 10 months ago

Oh I get it now

simon maxwell - 4 years, 10 months ago

Gotffried Leibniz!

Noel Cruz - 4 years, 5 months ago

hey... i don't know nuthin but dzeta-function of 2 is pi*pi/6... then the same function of 4 is (pi^4)/90.... AND!!! the same function of 3 up to this day is (as long as i remember) JUST PROVEN TO BE IRRATIONAL!! nothing more!!!! have i really missed some news?? if you found that thing of yours to be equal to 1 then you automatically have found dzeta(3), haven't you? or am i missing something here?

Nik Gibson - 2 years, 11 months ago

I did not realise the question was including all the next lines as well, so all my GPs had just 3 terms

Ken Garner - 4 months ago

The answer must be 3/4 I think??? I'm not sure...

Zachary Bisenio - 6 years, 1 month ago

Log in to reply

I'am also getting Same... But I realised both of us doing mistake..

Karan Shekhawat - 6 years, 1 month ago
Prasun Biswas
Apr 23, 2015

Here's another way to present your solution (it uses the same idea as that of your solution):

S = ∑ k = 2 ∞ ( ζ ( k ) − 1 ) = ∑ j = 2 ∞ ∑ i = 2 ∞ 1 j i = ∑ j = 2 ∞ 1 / j 2 1 − 1 / j = ∑ j = 2 ∞ j j 2 ( j − 1 ) = ∑ j = 2 ∞ ( 1 j − 1 − 1 j ) S=\sum_{k=2}^\infty(\zeta(k)-1)=\sum_{j=2}^\infty\sum_{i=2}^\infty \frac{1}{j^i}=\sum_{j=2}^\infty\frac{1/j^2}{1-1/j}=\sum_{j=2}^\infty\frac{j}{j^2(j-1)}=\sum_{j=2}^\infty\left(\frac{1}{j-1}-\frac{1}{j}\right)

By telescopy, the sum evaluates to,

S = 1 + ∑ j = 2 ∞ ( − 1 j + 1 j ) = 1 + ∑ j = 2 ∞ ( 0 ) = 1 S=1+\sum_{j=2}^\infty\left(-\frac{1}{j}+\frac{1}{j}\right)=1+\sum_{j=2}^\infty(0)=1

19 Helpful 2 Interesting 1 Brilliant 1 Confused

@Prasun Biswas , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 6 years, 1 month ago

Note the interchanging of the order of summations we do when we evaluate the double sum. That is to say, taking the sum given in question exactly in that order, we have,

S = ∑ k = 2 ∞ ( ζ ( k ) − 1 ) = ∑ i = 2 ∞ ∑ j = 2 ∞ 1 j i = ∑ j = 2 ∞ ∑ i = 2 ∞ 1 j i = 1 S=\sum_{k=2}^\infty(\zeta(k)-1)=\sum_{\color{#D61F06}{i=2}}^\infty\sum_{\color{#3D99F6}{j=2}}^\infty\frac{1}{j^i}=\sum_{\color{#3D99F6}{j=2}}^\infty\sum_{\color{#D61F06}{i=2}}^\infty\frac{1}{j^i}=1

The interchanging of the order of summation here is allowed by a special case of Tonelli's theorem since 1 j i \dfrac{1}{j^i} is non-negative ∀ 2 ≤ i , j \forall~2\leq i,j .

Prasun Biswas - 6 years, 1 month ago
William Chau
Apr 24, 2015

The value of the series is

(1/2^2+1/2^3+1/2^4+...)+(1/3^2+1/3^3+1/3^4+...)+(1/4^2+1/4^3+1/4^4+...)+...

= (1/2^2)(1+1/2+1/2^2+...)+(1/3^2)(1+1/3+1/3^2+...)+(1/4^2)(1+1/4+1/4^2+...)+...

= (1/2^2)/(1-1/2)+(1/3^2)/(1-1/3)+(1/4^2)/(1-1/4)+...

= (1/2^2)(2)+(1/3^2)(3/2)+(1/4^2)(4/3)+...

= 1/(1 * 2)+1/(2 * 3)+1/(3 * 4)+...

= 1-1/2+1/2-1/3+1/3-1/4+...

= 1.

18 Helpful 0 Interesting 0 Brilliant 0 Confused
Ervyn Manuyag
Nov 20, 2018

If you calculate it in ur head then the answer is 1

1 Helpful 0 Interesting 4 Brilliant 2 Confused
Bandit Eastern
Jul 13, 2018

dude I don't know any calculus and I guessed this one on the first try :/

0 Helpful 0 Interesting 2 Brilliant 1 Confused

Impressive solution

Valentin Duringer - 9 months ago
Mohammad Khaza
Jun 30, 2017

the value is,

(1/2^2+1/2^3+1/2^4+...)+(1/3^2+1/3^3+1/3^4+...)+(1/4^2+1/4^3+1/4^4+...)+.....................

= (1/2^2)(1+1/2+1/2^2+...)+(1/3^2)(1+1/3+1/3^2+...)+(1/4^2)(1+1/4+1/4^2+...)+................

= (1/2^2)/(1-1/2)+(1/3^2)/(1-1/3)+(1/4^2)/(1-1/4)+...............

= (1/2^2)(2)+(1/3^2)(3/2)+(1/4^2)(4/3)+.................

= 1/(1 * 2)+1/(2 * 3)+1/(3 * 4)+....................

= 1-1/2+1/2-1/3+1/3-1/4+...

= 1.

1 Helpful 0 Interesting 0 Brilliant 1 Confused
Julian Branker
Apr 25, 2015

This is right from Euler and his infinite series

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Brilliant solution

Valentin Duringer - 9 months ago
Jeff Giff
May 31, 2020

Note that
( 1 2 1 (\frac{1}{2^{1}} + 1 2 2 . . . ) \frac{1}{2^{2}}...) + ( 1 3 1 (\frac{1}{3^{1}} + 1 3 2 . . . ) \frac{1}{3^{2}}...) +...
= 1 1 \frac{1}{1} + 1 2 \frac{1}{2} + 1 3 \frac{1}{3} ...
And the difference between the equation above and the question is:
1 2 1 \frac{1}{2^{1}} + 1 3 1 \frac{1}{3^{1}} + 1 4 1 \frac{1}{4^{1}} +….
So the equation in the question equals to
( 1 1 (\frac{1}{1} + 1 2 \frac{1}{2} + 1 3 + … ) \frac{1}{3}+…) - ( 1 2 (\frac{1}{2} + 1 3 \frac{1}{3} + 1 4 + … ) \frac{1}{4}+…)
= 1 \boxed{1}






0 Helpful 0 Interesting 0 Brilliant 0 Confused
Otto Bretscher
Apr 24, 2015

One can easily create some interesting variants of this basic problem, things like ∑ s = 1 ∞ ( ζ ( 2 s ) − 1 ) , ∑ s = 2 ∞ ∑ n = 1 ∞ 1 ( 2 n ) s , \sum_{s=1}^{\infty}(\zeta(2s)-1), \quad\quad \sum_{s=2}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(2n)^s}, ∑ s = 1 ∞ ∑ n = 1 ∞ 1 ( 4 n ) 2 s , . . . \sum_{s=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(4n)^{2s}},\quad...

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Try to guess HAHAHAHA

0 Helpful 0 Interesting 0 Brilliant 2 Confused

This problem is interesting in it's own right, and I enjoyed reading the solution. However, I wouldn't call this a "Level 1" understanding of Geometric Series by any stretch of the imagination. It is frustrating to see so many questions in these quizzes so grossly miscategorized.

Michael Malione - 2 years, 2 months ago

Try this way Add all the Open all the brackets 1/2 series in gp 1/3 series in gp 1/4 series in gp so on Then It will be like 1/2 +1/3 + 1/4 ..... so on Infinite term of an hp So the answer is infinite

Abhishek dhok - 3 months, 1 week ago

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...