How Does the Sequence Convergence to 1/2 Work?

[quasar987]

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

\left\{\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right\}

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]

Is there a way to find this analytically?

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

 

[quasar987]

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

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: https://www.physicsforums.com/showthread.php?t=47964

 

[quantumdude]

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

\left\{\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}\right\}

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

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.

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?

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

 

[shmoe]

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.

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.

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

It's the limit of a_n 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 b_1=a_1, b_n=a_n-a_{n-1}, then a_n=\sum_{i=1}^{n}b_i, but that can be an awkward thing to do. In this case we'd find b_n=\frac{1}{2n(n+1)}, but I don't think that's what you were getting at?

 

[quantumdude]

I really do know better than that...

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