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our paper contains a direct elementary proof of the following series transformation :    ( * landen s transformation * ) if @xmath0 , then@xmath1{x}}{1+\sqrt{x}}\right)^{2}-\frac{1^{2}\cdot 3}{2^{2}\cdot 4^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{4}-\frac{1^{2}\cdot 3^{2}\cdot 5}{2^{2}\cdot 4^{2}\cdot 6^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{6}-\cdots\right\}=      & \\      & \\={\displaystyle}1+\left(\frac{1}{2}\right)^{2}x+\left(\frac{1\cdot 1}{2\cdot 4}\right)^{2}x^{2}+\left(\frac{1\cdot 1\cdot 3}{2\cdot 4\cdot 6}\right)^{2}x^{3}+\left(\frac{1\cdot 1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}\right)^{2}x^{4}+\cdots &   \\\end{array } $ } \ ] ]    @xmath2    the interest in this theorem is that it gives a direct proof the equality of the following two formulas for the perimeter of an ellipse with semi - axes @xmath3 and @xmath4 . ( * the perimeter of an ellipse * ) the perimeter , @xmath5 , of an ellipse with semi - major axes @xmath3 and @xmath4 and eccentricity @xmath6 is:@xmath7@xmath2    the first formula for @xmath5 is due to maclaurin @xcite in 1742 while the second is due to ivory @xcite in 1796 . berndt ( @xcite , p. 147 ) points out that ivory s formula can be obtained from a form of landen s transformation to be found in erdlyi s compendium ( @xcite , p.111 , formula ( 5 ) ) , but we have not been able to encounter a direct elementary proof in the literature which transforms the first of the two formulas for @xmath5 into the second one , and which does not appeal to general transformation formulas . ivory s own proof cleverly finesses this difficulty by showing that both series are equal to an integral . the following identity is valid for all integers @xmath8 and real numbers @xmath9.@xmath10    the following identities between binomial coefficients are well known:@xmath11 for example , see knuth ( @xcite pp . 71 and 46 , resp . ) now solve the first identity for @xmath12 , write @xmath13 in place of @xmath9 everywhere , use the second identity to get rid of the negative sign inside the binomial coefficients , and finally multiply and divide by @xmath14    this next lemma is the novelty in our proof . the following relation is valid for all integers @xmath15 @xmath16    we construct , in _ two different ways _ , a certain polynomial , @xmath17 , whose coefficient of @xmath18 is the sum @xmath19 defined in ( 3 ) . observe that the _ coefficient of _ @xmath20 in @xmath21is equal to ( put @xmath22)@xmath23therefore , _ the coefficient of _ @xmath24in@xmath25is equal to@xmath26thus , _ the sum _ we are seeking to evaluae , viz.@xmath27is the _ coefficient _ of @xmath18 in the _ _ sum__@xmath28this is the * _ first form _ * of the polynomial @xmath29    but @xmath17 can also be written@xmath30which is the * _ second form _ * of @xmath17 . expanding it by the binomial theorem we obtain@xmath31we see that all the powers of @xmath32 are _ even_. thus , if @xmath9 is _ odd _ , the coefficient of @xmath18 is _ zero _ , while if it is _ even _ , the coefficient of @xmath33is equal to@xmath34    the following result is well known ( @xcite ) :    ( * vandermonde s theorem * ) the following relation holds for all real numbers @xmath3 and @xmath4:@xmath35    @xmath2 our proof of * theorem 1 * is based on an idea of carlson @xcite . however we use the fundamental * lemma 2 * , of which we have given a direct elementary proof based on the elementary properties of the binomial coefficients ( see  2 ) , while carlson proves the transformation on the basis of general expansion theorems for  r - functions . " our proof consists of two steps :      the left - hand side of * theorem 1 * is rearranged into the following power series in @xmath36:@xmath37we achieve this via the binomial theorem , * lemma 1 * and * lemma 2*.      the power series @xmath38 is rearranged into a power series in @xmath32 alone via the binomial theorem . applying * lemma 1 * and vandermonde s theorem , the coefficients of this series collapse down to the coefficients of the right hand side of * theorem 1*. all subsequent series converge absolutely and uniformly for @xmath0 , so that the following series manipulations are valid in that range.@xmath39{x}}{1+\sqrt{x}}\right)^{2}-\frac{1^{2}\cdot 3}{2^{2}\cdot 4^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{4}-\frac{1^{2}\cdot 3^{2}\cdot 5}{2^{2}\cdot 4^{2}\cdot 6^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{6}-\cdots\right\}\\      & = { \displaystyle}(1+\sqrt{x})\sum_{m=0}^{\infty}\frac{-1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}\left\{{\displaystyle}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{2}\right\}^{m}\\ & = { \displaystyle}(1+\sqrt{x})\sum_{m=0}^{\infty}\frac{-1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}{\displaystyle}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{2 m }    \\      & = { \displaystyle}\sum_{m=0}^{\infty}\frac{-1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}4^{m}(\sqrt{x})^{m}(1+\sqrt{x})^{-2m+1 } \\   & = { \displaystyle}\sum_{m=0}^{\infty}\frac{-1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}4^{m}(\sqrt{x})^{m}(1 + 2\sqrt{x}+x)^{-m+\frac{1}{2}}\\ & = { \displaystyle}\sum_{m=0}^{\infty}\frac{-1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}4^{m}(\sqrt{x})^{m}(1+x)^{\frac{1}{2}}(1+x)^{-m}\left(1+\frac{2\sqrt{x}}{1+x}\right)^{-m+\frac{1}{2}}\\ & = -\sum_{m=0}^{\infty}\frac{1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}4^{m}\left(\frac{\sqrt{x}}{1+x}\right)^{m}(1+x)^{\frac{1}{2}}\sum_{k=0}^{\infty}\binom{-m+\frac{1}{2}}{k}\left(\frac{2\sqrt{x}}{1+x}\right)^{k}\\ & = -(1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\frac{1}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}2^{m}\sum_{k=0}^{\infty}\binom{-m+\frac{1}{2}}{k}\left(\frac{2\sqrt{x}}{1+x}\right)^{k+m}\\ & = -(1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\sum_{k=0}^{\infty}\frac{2^{m}}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}\binom{-m+\frac{1}{2}}{k}\left(\frac{2\sqrt{x}}{1+x}\right)^{k+m}\\ & = -(1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\sum_{k=0}^{\infty}\frac{2^{m}}{2m-1}\left\{\frac{1}{4^{m}}\binom{2m}{m}\right\}^{2}\frac{2m-1}{2m-1 + 2k}\binom{-m-\frac{1}{2}}{k}\left(\frac{2\sqrt{x}}{1+x}\right)^{k+m}\\ & \left(\text{since $ \binom{r}{k}=\dfrac{r}{r - k}\binom{r-1}{k}$}\right)\\ & = -(1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\sum_{k=0}^{\infty}\frac{2^{m}}{4^{2m}}\binom{2m}{m}^{2}\frac{1}{2m+2k-1}\frac{{\displaystyle}\binom{2m+2k}{m+k}\binom{m+k}{k}}{{\displaystyle}\binom{2m}{m}}\frac{(-1)^{k}}{4^{k}}\left(\frac{2\sqrt{x}}{1+x}\right)^{k+m}\\ & ( \text{by \textbf{lemma 1}})\\ & = ( 1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\sum_{k=0}^{\infty}\frac{2^{m}}{4^{2m}}\binom{2m}{m}\frac{1}{2m+2k-1}{\displaystyle}\binom{2m+2k}{m+k}\binom{m+k}{k}{\displaystyle}\binom{2m}{m}\frac{(-1)^{k}}{4^{k}}\left(\frac{2\sqrt{x}}{1+x}\right)^{k+m}\\ & ( \text{now put $ k:=n - m$ and interchange the order of summation})\end{aligned}\ ] ]    @xmath40    @xmath41 \left(\frac{2\sqrt{x}}{1+x}\right)^{n } ( \text{by \textbf{lemma 2}})\\   & = ( 1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\frac{-1}{4m-1}\frac{1}{4^{2m}}\binom{4m}{2m}{\displaystyle}\frac{1}{4^{m}}\binom{2m}{m}\left(\frac{2\sqrt{x}}{1+x}\right)^{2 m } ( \text{putting $ n=2m$}.)\end{aligned}\ ] ]    putting together the two extreme members of this chain of equalities , * _ we have shown : _ *    @xmath42{x}}{1+\sqrt{x}}\right)^{2}-\frac{1^{2}\cdot 3}{2^{2}\cdot 4^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{4}-\frac{1^{2}\cdot 3^{2}\cdot 5}{2^{2}\cdot 4^{2}\cdot 6^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{6}-\cdots\right\}\\         & = ( 1+x)^{\frac{1}{2}}\sum_{m=0}^{\infty}\frac{-1}{4m-1}\frac{1}{4^{2m}}\binom{4m}{2m}{\displaystyle}\frac{1}{4^{2m}}\binom{2m}{m}\left(\frac{2\sqrt{x}}{1+x}\right)^{2m}.   \end{aligned}\ ] ]    @xmath2    this completes the proof of * step 1*. we expand the term @xmath43 by the binomial theorem . therefore , the rhs of the equation in * theorem 3 * becomes@xmath44again , we put @xmath45 , interchange the order of summation , and observe that * _ the coefficient of _ * @xmath46 * * _ is_**@xmath47where we have used both the conclusion and the proof of * lemma 1*. so , if we now substitute this last expression:@xmath48into the rhs of ( 5 ) we obtain@xmath49{x}}{1+\sqrt{x}}\right)^{2}-\frac{1^{2}\cdot 3}{2^{2}\cdot 4^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{4}-\frac{1^{2}\cdot 3^{2}\cdot 5}{2^{2}\cdot 4^{2}\cdot 6^{2}}\left(\frac{2\sqrt[4]{x}}{1+\sqrt{x}}\right)^{6}-\cdots\right\}\\      & = \sum_{n=0}^{\infty}\sum_{m=0}^{n}\frac{(-1)^{n-1}}{2n-1}\frac{1}{4^{n}}\binom{2n}{n}\binom{-n+\frac{1}{2}}{m}\binom{n}{n - m}x^{n}\\ & = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{2n-1}\frac{1}{4^{n}}\binom{2n}{n}\left\{\sum_{m=0}^{n}\binom{-n+\frac{1}{2}}{m}\binom{n}{n - m}\right\}x^{n}\\ & ( \text{and by \textsc{vandermonde } 's theorem this is equal to } )       \end{aligned}\ ] ]
we prove landen s transformation by direct elementary transformation of infinite series .