Convergence of a sequence

quasar987

Apparently (according to my textbook), the sequence defined by

[tex]\left\{\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right\}[/tex]

converges towards 1/2, i.e. has 1/2 as a limit.

How could that be?! It seems to me that as n approaches infinity, all the fractions fall to zero. What is it I'm missing?

Hurkyl

The question is if they tend to zero faster than their number grow towards infinity.

quasar987

Is there a way to find this analytically?

arildno

Sure; you may write the partial sum as:
[tex]\frac{1}{n^{2}}(1+2+++n-1)=\frac{1}{n^{2}}\frac{n(n-1)}{2}[/tex]

quasar987

Gauss's sum again! Damn! You guys are smart, are you all doctors in mathematics or physics?

Manchot

Some of them are. :) Don't worry about it, I feel the same way you do all the time.

quasar987

I posted a new thread in "College Level Help" by mistake: http://www.physicsforums.com/showthread.php?t=47964

Tom Mattson

Hold on a second. How is it that the index appears in every term when you list out the series?

Also, the above is a series, not a sequence.

The limit of the sequence is zero.
The limit of the sequence of partial sums is 1/2.

shmoe

Tom, it's precisely the fact that the index "n" appears in each of the terms that makes this a sequence, and not a series, as it's given.

[tex]a_n=\sum_{i=1}^{n-1}\frac{i}{n^2}[/tex]

It's the limit of [tex]a_n[/tex] he's after. Since each of the terms in the sum is dependant on n, you can't break it into a series as I suspect you are thinking of doing.


You can of course think of any sequence as a series, by setting [tex]b_1=a_1, b_n=a_n-a_{n-1}[/tex], then [tex]a_n=\sum_{i=1}^{n}b_i[/tex], but that can be an awkward thing to do. In this case we'd find [tex]b_n=\frac{1}{2n(n+1)}[/tex], but I don't think that's what you were getting at?

Tom Mattson

Do me a favor and just ignore me for the rest of the night....