- Thread starter
- #1

#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

In the Math Challenge Forum it has been requested fo compute the series...

$\displaystyle S = \sum_{n=1}^{\infty} \tan^{-1}\ \frac{\sqrt{3}}{n^{2} + n + 3}\ (1)$

... and that has been performed using the general identity...

$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\ \frac{c}{n^{2} + n + c^{2}} = \tan^{- 1} c\ (2)$

... so that is $\displaystyle S = \tan^{- 1} \sqrt{3} = \frac{\pi}{3}$. Scope of this thread is to find a general procedure to construct series of inverse functions.

Let be $f(*)$ a strictly increasing function that maps an open inteval $[\alpha,\beta]$ containing 0 onto an interval $[a,b]$. We know that is such conditions the inverse function $f^{-1} (*)$ exists and it maps $[a,b]$ onto $[\alpha,\beta]$. Lets define $f(0) = s$.

Now we assume that $f(*)$ is solution of a functional equation...

$\displaystyle f(x - y) = G \{f(x),f(y)\}\ (3)$

Setting $f(x)=u$ and $f(y)=v$ the the so called

$\displaystyle f^{-1} (u) - f^{- 1} (v) = f^{-1} \{G(u,v)\}\ (4)$

We can use (4) to construct a telescopic series. If $\displaystyle c_{n},\ n=0,1,...$ is an increasing sequence in $[a,b]$ converging to $c \in [a,b]$, then...

$\displaystyle \sum_{k=1}^{n} \{f^{-1} (c_{k}) - f^{-1} (c_{k-1})\} = f^{-1} (c_{n})\ (5)$

... and that means that...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k},c_{k-1})\} = f^{-1} (c) - f^{- 1} (c_{0})\ (6)$

If $\displaystyle c_{n}$ is a strictly decreasing sequence converging to c the result is similar...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k-1},c_{k})\} = f^{-1} (c) + f^{- 1} (c_{0})\ (7)$

In next posts we will illustrate some interesting examples of use of (6) and (7)...

Kind regards

$\chi$ $\sigma$

$\displaystyle S = \sum_{n=1}^{\infty} \tan^{-1}\ \frac{\sqrt{3}}{n^{2} + n + 3}\ (1)$

... and that has been performed using the general identity...

$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\ \frac{c}{n^{2} + n + c^{2}} = \tan^{- 1} c\ (2)$

... so that is $\displaystyle S = \tan^{- 1} \sqrt{3} = \frac{\pi}{3}$. Scope of this thread is to find a general procedure to construct series of inverse functions.

Let be $f(*)$ a strictly increasing function that maps an open inteval $[\alpha,\beta]$ containing 0 onto an interval $[a,b]$. We know that is such conditions the inverse function $f^{-1} (*)$ exists and it maps $[a,b]$ onto $[\alpha,\beta]$. Lets define $f(0) = s$.

Now we assume that $f(*)$ is solution of a functional equation...

$\displaystyle f(x - y) = G \{f(x),f(y)\}\ (3)$

Setting $f(x)=u$ and $f(y)=v$ the the so called

*subtaction formula*permits us to write...$\displaystyle f^{-1} (u) - f^{- 1} (v) = f^{-1} \{G(u,v)\}\ (4)$

We can use (4) to construct a telescopic series. If $\displaystyle c_{n},\ n=0,1,...$ is an increasing sequence in $[a,b]$ converging to $c \in [a,b]$, then...

$\displaystyle \sum_{k=1}^{n} \{f^{-1} (c_{k}) - f^{-1} (c_{k-1})\} = f^{-1} (c_{n})\ (5)$

... and that means that...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k},c_{k-1})\} = f^{-1} (c) - f^{- 1} (c_{0})\ (6)$

If $\displaystyle c_{n}$ is a strictly decreasing sequence converging to c the result is similar...

$\displaystyle \sum_{k=1}^{\infty} f^{-1} \{G(c_{k-1},c_{k})\} = f^{-1} (c) + f^{- 1} (c_{0})\ (7)$

In next posts we will illustrate some interesting examples of use of (6) and (7)...

Kind regards

$\chi$ $\sigma$

Last edited: