The book I am using for my Advance Calculus course is Introduction to Analysis by Arthur Mattuck.

$\bf 1.$ By mimicking the argument in $4.2$, prove Leibniz' famous result $$\frac\pi4=\lim_{n\to\infty}1-\frac13+\frac15-\frac17+\ldots+\frac{(-1)^{n}}{2n+1}:$$ $(a)$ derive a formula $1-u^2+u^4-\ldots+(-1)^{n}u^{2n}=\frac1{1+u^2}+e_n(u)$, where the error term $e_n(u)$ depends on $u$;
$(b)$ integrate every term from $u=0$ to $u=1$, and show the integral of the error term tends to $0$ as $n\to\infty$. (Compare it with another definite integral.)

Proposition $\bf 4.2$ The geometric sum limit. Consider the geometric sum $$a_n=1+a+a^2+\ldots+a^n;\tag{3}$$ then $$\lim_{n\to\infty}a_n=\frac1{1-a},\quad\text{if }|a|<1.$$ Proof. The compact formula for $a_n$ is (check it by cross multiplying): $$1+a+a^2+\ldots+a^n=\frac{1-a^{n+1}}{1-a}=\frac1{1-a}-\frac{a^{n+1}}{1-a},\ a\neq 1.$$ Comparing $(4)$ with $(1)$, we see that the error term for the desired limit is $$e_n=\frac{-a^{n+1}}{1-a},\quad a\neq 1.$$ By Theorem $3.4$, if $|a|<1$, then $a^n\to0$. It follows that $e_n\to 0$ (see Question $3.2/2$); by the error-form principle, this proves Proposition $4.2$.
We can make similar limit proofs and study convergence rate for sequences related to $(3)$. Here is the one that started off this chapter.

This is my rough proof to this question. I was wondering if anybody can look over it and see if I made a mistake or if there is a simpler way of doing this problem. I want to thank you ahead of time it is greatly appreciated.So lets begin:


enter image description here enter image description here


I think the simplest way to arrive at this conclusion is to use Gregory's series (basically a Taylor series for $\arctan(x)$... see this link; we get it the same way we get the Taylor series for the $\sin$ function). From Gregory's series we get $$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} +...$$ Now we just plug 1 into the series above and you get the formula you were proving. This is not necessarily the proof the book asks for, but is by far the simplest in my opinion.

Note: If you go to the Wikipedia link I attach a more complete proof using integrals is shown without showing Gregory's series prior, and the page also contains a link to Leibniz's original geometric proof.

  • 1
    $\begingroup$ But this approach needs Abel's theorem to guarantee that the value you get when you plug in $x=1$ is actually $\arctan 1$ (since we are on the boundary of the disc of convergence for the power series). $\endgroup$ – mrf Sep 24 '15 at 5:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.