IIT JEE , AIEEE Forum - Mathematics , Algebra - again an open challenge to all community members and experts

x2008

solve 1 + 2 + 3 + 4 + ….∞

pantpranav

Sorry this is not an equation, so I cannot solve it.

But the expression has the value: n(n+1)/2

rudra.panda

it is infinity only.

krish1092

it is infinity

most.happening

what a dumb challenge..................logically infinite

hsbhatt

The answer is $\boxed{-\frac{1}{12}}$ by Ramanujan Summation.

Check out http://en.wikipedia.org/wiki/Divergent_series The foot of the article mentions summation of 1+2+3+...

Also see http://www.goiit.com/posts/list/differenciation-so-you-thought-you-were-good-at-numbers-44321.htm for a lively discussion we had earlier on this matter.

Entertainment Guaranteed!

Moral: Never jump to hasty conclusions about people

rudra.panda

Consider the sum $-\sum_{i=1}^\infty i(-r)^i$ . Let this sum equal S. Then:

$S=r-2r^2+3r^3-4r^4+...$

$Sr=0+r^2-2r^3+3r^4-...$

$S+Sr=r-r^2+r^3-r^4+...$

$S=\frac r{(r+1)^2}$

If we take the limit as r approaches 1, $1-2+3-4+...=\frac 14$

Now, let $1+2+3+...=T$ . Then $2(2+4+6+8+...)=4T$ and $\frac 14+4T=T$ , so $\frac 14=-3T$ and

$1+2+3+4+...=-\frac 1{12}$

Quite an interesting topic......................

rudra.panda

sorry forgot to mention the source. http://www.mathlinks.ro/viewtopic.php?t=78489

rudra.panda

Have a look at this:

$\frac{f\left( 0\right) }{2}+f\left( 1\right) +\cdots+f\left( n-1\right) +<br/>\frac{f\left( n\right) }{2}<br/>=\frac{f\left( 0\right) +f\left( n\right) }{2}+\sum_{k=1}^{n-1}f\left(<br/>k\right) = \int_0^n f(x)\,dx + \sum_{k=1}^p\frac{B_{k+1}}{(k+1)!}\left(f^{(k)}(n)-f^{(k)}(0)\right)+R$

or simply:

$f\left( 1\right)+f\left( 2\right) +\cdots+f\left( n-2\right) + f\left( n-1\right)= C + \int_1^n f(x)\,dx + \sum_{k=1}^\infty\frac{B_{k+1}}{(k+1)!}\left(f^{(k)}(n)-f^{(k)}(0)\right)$

Where C is a constant specific to the series and its analytic continuum. This he proposes to use as the sum of the divergent sequence. It is like a bridge between summation and integration. Using standard extensions for known divergent series, he calculated "Ramanujan summation" of those. In particular, the sum of 1 + 2 + 3 + 4 + · · · is

$1+2+3+\cdots = -\frac{1}{12} (\Re)$

(Wikipedia)

feynmann

Hey , Rudra would u plz kindly let us know that how have u taken the limit ?

The summation formula that u have used is valid only for r<1 ( and clearly the RHL as r->1+ does not exist !! )

GoNik

yaaaaaaaaaar the graph takes it to infinity....how could that be then??....the graph doesnt stop at any point....wat say??

GoNik

that limit that was applied went for r----->1.......how can u add to T..??.....they are 2 different expressions..correct me kindly if i am wrong....

GoNik

reply pleeeeeeeeez

Ask & Discuss Questions with Community & Experts