Does the Sequence a_n Converge or Diverge as n Approaches Infinity?

[ProBasket]

Determine whether the sequence a_n = \frac{1^2}{n^3} + \frac{2^2}{n^3} + ... + \frac{n^2}{n^3} converges or diverges. If it converges, find the limit.


wouldnt it converge to 0?

 

[Hurkyl]

Maybe, maybe not. Why do you think it will go to 0? Any ideas on how to go about trying to prove it?

 

[James R]

That seems like a difficult problem, since you have a sequence of series.

The first few terms in the sequence would be:

\left{1, \frac{1^2}{2^3} + \frac{2^2}{2^3}, \frac{1^2}{3^3} + \frac{2^2}{3^3} + \frac{3^2}{3^3}, \frac{1^2}{4^3} + \frac{2^2}{4^3} + \frac{3^2}{4^3} + \frac{4^2}{4^3},...\right}

 

[quasar987]

The nth term is increasing and bounded by one

a_n \leq \frac{n^2}{n^3} + \frac{n^2}{n^3} + ... + \frac{n^2}{n^3} = n\frac{n^2}{n^3} = 1

 

[quasar987]

Actually, the nth term can be rewritten as

a_n = \frac{1}{n^3}\sum_{i=1}^{n}i^2

Do you recognize this sum?

 

[quasar987]

The reason why the value of convergence of this sequence is not simply obtained by doing

\lim_{n\rightarrow \infty} \frac{1^2}{n^3} + \frac{2^2}{n^3} + ... + \frac{n^2}{n^3} = \lim_{n\rightarrow \infty} \frac{1^2}{n^3} + \lim_{n\rightarrow \infty} \frac{2^2}{n^3} + ... + \lim_{n\rightarrow \infty} \frac{n^2}{n^3} = 0+0+...+0

is that the NUMBER OF TERMS in the sum augment to infinity as well! So you can't be sure what the sum's going to be: even though all members of the sum go to zero, there is an infinity of them.

 

[ProBasket]

so it converges, but no answer?

 

[mugzieee]

just a thought, wouldn't the limit comparison test work on this if you choose your b_n to be 1/n, and use the p-series to say b_n diverges?

 

[Hurkyl]

Pro: we're here to help, not give answers. You should have enough information to figure it out yourself.

mug: what are you comparing to 1/n?