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For instance in my recent post:

I have this limit to find $$\lim_{n\to \infty }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots+\left(-1\right)^{n-1}\cdot \frac{1}{2n-1}\right)=\text{ ?}$$

and we know too this integral $$I_n=\int _0^1\:\frac{x^n}{x^2+1}dx$$ and that relation for recurrence: $$I_{2n}\:\cdot \:\left(-1\right)^{n-1}\:=\:\frac{\left(-1\right)^{n-1}}{2n-1}-\:\left(-1\right)^{n-1}\cdot I_{2n-2}$$

Okay and now how we can continue, if we know recurrence relation because my teacher adviced me to use this and I don't know what helps me, because if I put value for n>2, I'll find some terms and I need to find the sums...

okay so tell me someone if that recurrence relation can helps me and if not put on hold...

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  • $\begingroup$ Maybe you can use the elementary proof. $\endgroup$ Mar 29, 2015 at 2:03
  • $\begingroup$ what helps me a recurrence relation if I find it? that I want to know... my teacher advised me to find a recurrence relation ! why? because I have sum and if I find a recurrence relation just will know some terms not all and not sum of them, and is not help... just I think, maybe I am wrong... $\endgroup$
    – Lucas
    Mar 29, 2015 at 2:05
  • $\begingroup$ @Jan-MagnusØkland That was my first impulse too, but OP wants to avoid using Taylor series, and all three of the proofs at the link use the Taylor series of $\arctan x$ at $0$. $\endgroup$ Mar 29, 2015 at 2:06
  • $\begingroup$ Can you use integration? $\endgroup$ Mar 29, 2015 at 2:16
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    $\begingroup$ Let $$S_{n} = 1 - \frac{1}{3} + \cdots + (-1)^{n - 1}\frac{1}{2n - 1}$$ and we add your original recurrence relations for $n = 1, 2, \ldots, n$ we get $(-1)^{n - 1}I_{2n} = S_{n} - I_{0} = S_{n} - (\pi/4)$. To evaluate limit of $S_{n}$ just show that $I_{2n} \to 0$ as $n \to \infty$. This is easily seen if we note that $$0 \leq I_{2n} = \int_{0}^{1}\frac{x^{2n}}{1 + x^{2}}\,dx \leq \int_{0}^{1}x^{2n}\,dx = \frac{1}{2n + 1}$$ and apply Squeeze theorem. $\endgroup$
    – Paramanand Singh
    Mar 31, 2015 at 12:37

2 Answers 2

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(Not sure if this is the method your teacher is looking for. Probably not...)

You're looking for the infinite sum:$\renewcommand{\d}[1]{\operatorname{d}\!{#1}}$ $$1-\frac13+\frac15-\frac17+\dotsb$$

Now, note that: $$\int_0^1t^{n-1}\d t=\left[\frac{t^n}n\right]_0^1=\frac1n$$ This is central to how I'm going to solve this problem. In particular, letting $n=0,2,4,$ etc., this gives us the family of equalities: $\int_0^1\d t=1,\int_0^1t^2\d t=\frac13,\int_0^1t^4\d t=\frac15,$ etc.

Thus, we have: \begin{align} 1-\frac13+\frac15-\frac17+\dotsb&=\int_0^1\d t-\int_0^1t^2\d t+\int_0^1t^4\d t-\dotsb\\ &=\int_0^1(1-t^2+t^4-\dotsb)\d t \end{align} You may recognize $1-t^2+t^4-\dotsb$ as a geometric series. (If you forgot: $1+u+u^2+\dotsb=\dfrac1{1-u}$.) Here we have $u=-t^2$, so we have:

$$1-t^2+t^4-\dotsb=\frac1{1+t^2}$$

Continuing, we have: \begin{align} 1-\frac13+\frac15-\frac17+\dotsb&=\int_0^1(1-t^2+t^4-\dotsb)\d t\\ &=\int_0^1\frac1{1+t^2}\d t\\ &=\left[\arctan t\right]_0^1\\ &=\arctan1-\arctan0\\ &=\frac\pi4-0\\ &=\frac\pi4 \end{align}

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  • $\begingroup$ Yes, I know that method... but we need to use with that recurrence relation who I found some minutes ago $\endgroup$
    – Lucas
    Mar 29, 2015 at 2:38
  • $\begingroup$ @Lucas Considering the fact that the sum is $\dfrac\pi4$... I really have no idea. $\endgroup$ Mar 29, 2015 at 2:39
  • $\begingroup$ @Lucas Well, I guess you can prove that $1-\dfrac12+\dotsb\dfrac1{2n+1}=\int_0^1\dfrac{1+t^n}{1+t^2}$, inductively, and then take the limit to get $\int_0^1\dfrac1{1+t^2}=\dfrac\pi4$. $\endgroup$ Mar 29, 2015 at 2:43
  • $\begingroup$ Who see my post please some help, because I don't understand why my teacher advised me to use this recurrence relation ... for what? $\endgroup$
    – Lucas
    Mar 29, 2015 at 2:45
  • $\begingroup$ columbus8myhw I can prove that, this was my first intuition too because I think is only method, maybe I will find new reviews later from other users, think so... $\endgroup$
    – Lucas
    Mar 29, 2015 at 2:52
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I'm going to use the fact that the series:

$\displaystyle 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\left(-1\right)^{n-1}\cdot \frac{1}{2n-1}$

is actually taylor series expansion of $tan^{-1}(x)$ at $x= 1$.

The function $tan^{-1}(x)$ satisfies the differential equation

$\displaystyle (x^{2}+1)\frac{d^2y(x)}{dx^2}+ 2x\frac{dy(x)}{dx} = 0$

with $y(0)=0, y^{'}(0)=1$.

Finding a recurrence for taylor coefficients $u(n)$ of $y(x)$ must solve your case now.

From initial conditions of differential equations, we can say $u(0) = 0, u(1) = 1$. Replacing $y(x)$ by $\displaystyle\sum_{n=0}^\infty u(n) x^{n}$ in the differential equation, we get

$\displaystyle2x\sum_{n=1}^\infty nu(n) x^{n-1} + (x^2+1)\sum_{n=2}^\infty n(n-1)u(n) x^{n-2} = 0$.

By collecting powers of $x^n$ and shifting summation indices,

$\displaystyle\sum_{n=0}^\infty (2n + n(n-1))u(n) x^{n}+ \sum_{n=0}^\infty (n+1)(n+2)u(n+2) x^{n} = 0$

By comparing coefficients of $x^n$, we get this recurrence relation for $n \ge 0$:

$\displaystyle (n^2 + n)u(n) + (n^2 + 3n + 2)u(n+2) = 0 $

Cancelling $(n+1)$ we get, $nu(n) + (n+2)u(n+2) = 0$ .

So the recurrence relation must be

$nu(n) + (n+2)u(n+2) = 0$

with $u(0) = 0, u(1) = 1$.

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