# sum of a series

• Feb 4th 2008, 01:36 PM
aston
sum of a series
hi, i have a series 1-1/3+1/5-1/7+..., which i think is also called the Leibniz formula, i know that to find the nth term you can can use (-1)^n/(2n+1), but i need help trying to find a formula or a way to find the sum of the series for a certain number of terms i.e. the sum of the first 10 terms, can anyone help?
• Feb 4th 2008, 02:29 PM
Krizalid
Your serie is equal to $\frac\pi4.$

Consider $\arctan x = \int_0^x {\frac{1}
{{1 + u^2 }}\,du} .$
So,

$\arctan x = \int_0^x {\frac{1}
{{1 + u^2 }}\,du} = \int_0^x {\left\{ {\sum\limits_{k = 0}^\infty {( - 1)^k u^{2k} } } \right\}\,du} .$

Finally

$\arctan x = \sum\limits_{k = 0}^\infty {( - 1)^k \left\{ {\int_0^x {u^{2k} \,du} } \right\}} = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k + 1} }}
{{2k + 1}}} ,\,\left| x \right| \le 1.$

Now set $x=1$ and we're done.
• Feb 4th 2008, 07:25 PM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
Your serie is equal to $\frac\pi4.$

Consider $\arctan x = \int_0^x {\frac{1}
{{1 + u^2 }}\,du} .$
So,

$\arctan x = \int_0^x {\frac{1}
{{1 + u^2 }}\,du} = \int_0^x {\left\{ {\sum\limits_{k = 0}^\infty {( - 1)^k u^{2k} } } \right\}\,du} .$

Finally

$\arctan x = \sum\limits_{k = 0}^\infty {( - 1)^k \left\{ {\int_0^x {u^{2k} \,du} } \right\}} = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k + 1} }}
{{2k + 1}}} ,\,\left| x \right| \le 1.$

Now set $x=1$ and we're done.

You cannot do that. Because $\frac{1}{1+u^2} = \sum_{k\geq 0} (-1)^k u^{2k}$ only when $|u|<1$. What you are doing is letting $u=1$ and it does not converge there.
• Feb 4th 2008, 07:54 PM
ThePerfectHacker
Quote:

Originally Posted by aston
hi, i have a series 1-1/3+1/5-1/7+..., which i think is also called the Leibniz formula, i know that to find the nth term you can can use (-1)^n/(2n+1), but i need help trying to find a formula or a way to find the sum of the series for a certain number of terms i.e. the sum of the first 10 terms, can anyone help?

Proof 1: First note that $\frac1{1+x^2} = 1 - x^2 + x^4 -...$ for $|x|<1$. This tells us that (integrate both sides) $\tan^{-1} x = x - \frac{x^3}{3}+\frac{x^5}{5}-...$ for $|x|<1$. Now to justify what Krizalid did is to use Abel's Theorem which tells us that the series $f(x)$ is continous at $x=1$ because it converges there (by alternating series test). Thus, if $x_k\to 1$ is a sequence in $(-1,1)$ then $\lim ~ f(x_k) = f(\lim x_k)$ thus $f(1) = \lim ~ f(x_k) = \lim ~ \tan^{-1} (x_k) = \frac{\pi}{4}$.

Abel's Theorem: If $f(x) = \sum_{n=0}^{\infty}a_nx^n$ is a power series of radius of convergence of radius $0 such that the series converges at $x=R$ then $f$ is continous at $x=R$.
• Feb 5th 2008, 07:47 PM
ThePerfectHacker
Here is a more elementary proof.

Proof 2: For any $x$ we have $\sum_{k=0}^{n}(-1)^k x^{2k} = \frac{1 +(-1)^{n+1} x^{2n+1}}{1+x^2}$. Thus, $\int_0^1 \sum_{k=0}^n (-1)^k x^{2k} dx = \int_0^1 \frac{dx}{x^2+1} \pm \int_0^1 \frac{x^{2n+1}}{x^2+1} dx$ where $\pm$ depends on the parity of $n$. This tells us that $\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2k+1} = \tan^{-1} 1 \pm \int_0^1 \frac{x^{2n+1}}{x^2+1} dx$. Thus, $\left| \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2k+1} - \frac{\pi}{4}\right| = \int_0^1 \frac{x^{2n+1}}{x^2+1} dx$. Now $\frac{x^{2n+1}}{x^2+1} \leq x^{2n+1}$. This means, $\left| \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2k+1} - \frac{\pi}{4}\right| \leq \int_0^1 x^{2n+1} dx = \frac{1}{2n+2} \leq \frac{1}{n}$. This means for any $\epsilon > 0$ if we chose $N> \frac{1}{\epsilon}$ it will mean that $\sum_{k=0}^N \frac{(-1)^kx^{2k+1}}{2k+1}$ is within $\epsilon$ of $\frac{\pi}{4}$. And so, by definition, this series converges to $\frac{\pi}{4}$.