# Writing a sequence in closed form

• Apr 18th 2009, 12:01 PM
Writing a sequence in closed form
Consider the sequence $\left\{ {{a_n}} \right\}_{n = 1}^{ + \infty }$, where

${a_n} = \frac{1^2}{{{n^3}}} + \frac{2^2}{{{n^3}}} + \cdots + \frac{n^2}{{{n^3}}}$

I've written out enough terms to guess that this sequence converges at 1/3, but I'm supposed to confirm my conjecture by expressing ${{a_n}}
$
in closed form and calculating the limit.

I understand the method for finding the sum of

${a_n} = 1 + 2 + \cdots + n$, which is $\frac{{n(n + 1)}}{2}$ in closed form,

but I'm having trouble figuring out an equivalent equation for

${a_n} = {1^2} + {2^2} + \cdots + {n^2}$

Any help would be very much appreciated.
• Apr 18th 2009, 12:10 PM
Plato
$\sum\limits_{k = 1}^n {k^2 } = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}
{6}$
• Apr 18th 2009, 12:57 PM
Thank you, Plato, but how did you arrive at that conclusion?
• Apr 18th 2009, 01:39 PM
Plato
Quote:

Thank you, Plato, but how did you arrive at that conclusion?

That is just a widely known fact.
One finds it in any calculus textbook.
Or prove it by induction.
• Apr 18th 2009, 03:37 PM
Jester
Quote:

Thank you, Plato, but how did you arrive at that conclusion?

Here's a way. Consider

$\sum_{i=1}^n (i+1)^3 - i^3$

If you write out the terms

$\sum_{i=1}^n (i+1)^3 - i^3 = \left( 2^3-1^3\right) + \left( 3^3-2^3\right) + \left( 4^3-3^3\right) + \cdots + \left( (n+1)^3-n^3\right)$

and you see all the terms cancel except two so

$\sum_{i=1}^n (i+1)^3 - i^3 = (n+1)^3-1 = n^3 + 3n^2 + 3n$

Now expand the sum

$\sum_{i=1}^n (i+1)^3 - i^3 = \sum_{i=1}^n i^3 + 3i^2 + 3i + 1 - i^3 = 3\sum_{i=1}^n i^2 + \underbrace{3\sum_{i=1}^n i + \sum_{i=1}^n 1}_{\text{we know these}}$ $= 3\sum_{i=1}^n i^2 + \frac{3}{2} n (n+1) + n$

Now equate each part

$3\sum_{i=1}^n i^2 + \frac{3}{2} n (n+1) + n = n^3 + 3n^2 + 3n$

Solving for the sum gives, nothing up my sleeve, presto

$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$