[Guest]

Look at the sequence,

t_n = (1/n^3)*(1+2^2+...+n^2).

t_n =(1/n)*((1/n)^2+(2/n)^2+...+(n/n)^2)

What is the limit?

What could it be geometrically?

I am curious to hear of your interpretations.

I guess this is more like a math riddle than a puzzle.

[Guest]

1/3

[Guest]

limit is 0.

don't know the shape..

[Guest]

The limit is 1/3... although I have to admit that I used some computer help, I have a pic of the plot this function describes...

[Guest]

The limit is 1/3... although I have to admit that I used some computer help, I have a pic of the plot this function describes...

DAMN!! I can't upload anything...

[Guest]

WolframAlpha couldn't find a Limit to such function

http://www.wolframalpha.com/input/?i=lim_(n-%3Einfinity)+(y(n)+%3D+(1%2B2^2%2B...%2Bn^2)/n^3)

[Guest]

F(n) = 1/n³(1² + 2² + 3² + ... n²)

(1² + 2² + 3² + ... n²) = n(n+1)(2n+1)/6

Then F(n) = 1/n³(n(n+1)(2n+1)/6)

F(n) = (n+1)(2n+1)/6n²

F(n) = 1/3 + 1/2n + 1/6n²

As n tends to infinity, F(n) = 1/3

Shape of the curve would be like a ice hockey stick

[Guest]

A Graph of what it looks like with solution:

[Guest]

jagdmc, you can use type sum n^2 in WolframAlpha, that it should give you the partial sum formula you can incorporate into your solution.

Given a prism or a cylinder, the volume of the pyramid of cone, with the same base spanning the same height will have volume of 1/3 that of the prism or cylinder containing it. It comes from the ever finer approximation of the pyramid and cone with smaller prisms, and cylinders.

In the other form from the riddle spoiler, the first part of the multiplication is 1/n or the scale of the height of the approximating disks (by n disks) to that of the whole prism, and the sum of the k/n squared gives the sum of the scaled area of the base of the n disks compared to the base. Taking the limit gives 1/3 which is that of the volume of the pyramid or cone with respect to that of its containing prism or cylinder.

[Guest]

Think of a pyramid made of cubes. The Bottom layer has nXn cubes, the next layer has (n-1)X(n-1) cubes, so on. At the top there's only one. The number of cubes needed for this pyramid is about 1/3 n^3 as n grows sufficiently large.