According to Wikipedia, Ramanujan came up with the following series expansion of the exponential integral:
$$\operatorname{Ei}(x)=\gamma+\ln|x|+e^{x/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1}$$
My first instinct was to try and verify the derivative:
$$\frac{e^x}x\stackrel?=\frac1x+\frac12e^{x/2}\left(\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1}\right)+e^{x/2}\left(\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}\right)$$
and then perhaps apply Cauchy products? If so, then this doesn't seem very nice at all.
How can I show this equality?
Also as an aside, but below the formula on Wikipedia are some claimed bounds, which no matter how I look at it, are wrong, or if perhaps a typo, what should they be?
Best Answer
You can subtract $\frac{1}{x}$ and multiply by $xe^{-x/2}$ to avoid the Cauchy product and instead use the known series expansions of $e^x$. You would get $$e^{x/2} -e^{-x/2} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}$$
Write out the series expansion of the left hand side to get $$\sum_{n=0}^{\infty} \frac{x^n}{2^n n!} - \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n n!} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}$$
From here, you can just show that the $x^n$ terms are the same on both sides. It's clearly equal for $n = 0$ and $n = 1$ ($[x^0] = 0$ and $[x^1] = 1$). Other than that, you would need to show the relation $$\frac{1}{2^n n!} - \frac{(-1)^n}{2^n n!} \stackrel?= \frac{1}{2}\frac{\left(-1\right)^{n-2}}{\left(n-1\right)!2^{n-2}}\sum_{k=0}^{\lfloor (n-2)/2 \rfloor}\frac{1}{2k+1}+\frac{\left(-1\right)^{n-1}}{\left(n-1\right)!2^{n-1}}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}\frac{1}{2k+1}$$
Multiplying by $2^n n!$ and factoring the RHS, this simplifies to $$1 - (-1)^n \stackrel?= 2n\left(-1\right)^{n}\left( \sum_{k=0}^{\lfloor (n-2)/2 \rfloor}\frac{1}{2k+1}-\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}\frac{1}{2k+1}\right)$$
Splitting this up into even and odd cases makes it easier. If $n$ is even, then the LHS is $0$ and $\lfloor (n-2)/2 \rfloor = \lfloor (n-1)/2 \rfloor$, so the RHS is $0$ as well. If $n$ is odd, then the LHS is $2$. The RHS would be $$-2n\left( -\frac{1}{2\lfloor (n-1)/2 \rfloor + 1} \right) = -2n \cdot \frac{-1}{n} = 2$$
Therefore, the left hand side and the right hand side are both equal.