Return to Answer

Wow! I'm really surprised that professional mathematicians could not answer such a trivial question for so long. This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions: $$ {}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{z^{k}}{(c+1)_k}\\ =\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots $$ Just put $z=-1$, $c=n+1$, (You are qualified to do this, at least, right? Great job! I can see your MIT education really pays for itself.)

If you don't like special functions then consider https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula . The formula above follows by direct application of Euler's formula.

This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions: $$ {}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{z^{k}}{(c+1)_k}\\ =\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots $$ Just put $z=-1$, $c=n+1$.

If you don't like special functions then consider https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula . The formula above follows by direct application of Euler's formula.

Wow! I'm really surprised that professional mathematicians could not answer such a trivial question for so long. This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions: $$ {}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{z^{k}}{(c+1)_k}\\ =\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots $$ Just put $z=-1$, $c=n+1$, (You are qualified to do this, at least, right? Great job! I can see your MIT education really pays for itself.)

Wow! I'm really surprised that professional mathematicians could not answer such a trivial question for so long. This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions: $$ {}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{z^{k}}{(c+1)_k}\\ =\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots $$ Just put $z=-1$, $c=n+1$, (You are qualified to do this, at least, right? Great job! I can see your MIT education really pays for itself.)

If you don't like special functions then consider https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula . The formula above follows by direct application of Euler's formula.

Wow! I'm really surprised that professional mathematicians could not answer such a trivial question for so long. This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions: $$ {}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{(-z)^{k}}{(c+1)_k}\\ =\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots $$ Just put $z=-1$, $c=n+1$, (You are qualified to do this, at least, right? Great job! I can see your MIT education really pays for itself.)

Wow! I'm really surprised that professional mathematicians could not answer such a trivial question for so long. This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions: $$ {}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{z^{k}}{(c+1)_k}\\ =\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots $$ Just put $z=-1$, $c=n+1$, (You are qualified to do this, at least, right? Great job! I can see your MIT education really pays for itself.)