Computing a Series Sum: n2 + (r-1)2 to n2 + (n-1)2

[andyrk]

Check you statement of the problem in post #10 and see if there is any justification for taking the limit of $\alpha_n$ as $n \rightarrow \infty$

Does the problem ask the value of $\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}$?
Or does it ask the value of $\lim_{n \rightarrow \infty}(\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} )$ ?
Did you omit the limt?

See my edited message. What you quoted, I changed it just a moment ago.
And the problem asks for the value of $\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}$.

 

[Stephen Tashi]

I must go visit someone now. I'll get back to this in about 4 hours.

 

[andyrk]

I must go visit someone now. I'll get back to this in about 4 hours.

Sure. :)
Read #34 once you come back. It has been edited a lot.

The thing to understand from the question was that the summations were not equal. But when the same summations were converted to integrals by the means of a limit, then the integrals became equal. So the area when represented in the form of a summation is not equal. But when the same summations are converted into integrals then the area represented by the definite integral is equal. (For this question only, that is). Am I right?

 

[Stephen Tashi]

So if I use the equation $\alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) )$ and not $\alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )$, I get the answer of $\frac{\alpha_n}{\alpha} - \frac{\beta_n}{\beta}$ as $\frac{-2}{nπ}$ (i.e k, where k is some negative real number) and $\alpha_n - \beta_n$ comes out to be $\frac{-1}{2n}$. Is that correct?

Yes.

Also, the reason which I came up with for $α = β = π/4$. First of all
$\alpha = \lim_{n\rightarrow \infty} \alpha_n =>$ $\alpha = \int_a^b f(x) dx$
and
$\beta = \lim_{n\rightarrow \infty} \beta_n =>$ $\beta = \int_a^b f(x) dx$

You should write "=" instead of "=>".

When we evaluate the above integral, it comes out to be $π/4$. Now the question is, why is $α = β$? $α$ and $β$ are equal because their integrals come out to be the same

I'd prefer to say "because they result from taking limits of sums that approximate the same integral".

But why do the two integrals come out to be the same even though the limits are different? So question reduces down to that why is $\lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)$? The reason which I could think of as to why the limits are equal is because when $n→∞$, the difference in areas under the two curves for $\alpha_n$ and $\beta_n$ can be neglected and they can be said to be approximately equal.

Yes.

They are approximately equal because when we look closely, we realize that for $\alpha_n$, the area under the curve is measured from $x = \frac{1}{n}$ to $x = 1$. But for $\beta_n$, the area under the curve is measured from $x = \frac{0}{n}$ to $x = \frac{n-1}{n}$. This difference in area becomes negligible when n→∞ and so both the areas are measured from x = 0 to x = 1. This is because, for $n→∞$, $\frac{1}{n} → 0$ and $\frac{n-1}{n} → 1$.

That's a correct intuitive explanation. To prove formally that the areas are the same is harder, but that proof should haven been given in your text materials when you were introduced to definite integrals.

As a result, $α = β$. Now, even if we use the equation $\alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )$, we get the value of $\alpha_n - \beta_n$ as $\frac{-n}{1+n^2}$ and hence $\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-n/(1+n^2)}{π/4} = \frac{-4n}{π(1+n^2)}$( i.e $k$, where $k$ is some negative real number).

if we use the equation $\alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)$, just the value of $\alpha_n - \beta_n$ changes from $\frac{-1}{2n}$ to $\frac{-n}{1+n^2}$. The answer to the whole problem still remains the same, i.e, $\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}$ still comes out to be a number $k$ which is a negative real number, regardless of the equation used. So $\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}$ = $\frac{\alpha_n}{\alpha} - \frac{\beta_n}{\alpha}$ = $\frac{\alpha_n - \beta_n}{\alpha}$ = $\frac{-1/2n}{π/4} = \frac{-2}{nπ} = k$, (where $k$ is a negative real number).

It will get confusing to discuss the results from two different problems. If you change the definitions of $\alpha_n$ and $\beta_n$ then you might have to change the limits of integration and the value of the integral would be different.

So, all in all, if we use the equation $\alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) )$, then $\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-2}{nπ} = k$.

Yes.

And if we use the equation $\alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )$

Let's not worry about that!

 

[andyrk]

Thanks! Nice problem though.