We already know that the irrational number $\pi$ can be expressed in this way:
$ \pi =4-\frac{4}{2}+\frac{4}{5}+\cdots +\left( -1 \right) ^{n+1}\frac{4}{2n-1}+\cdots=\sum\limits_{n=1}^\infty\left( -1 \right) ^{n+1}\frac{4}{2n-1} $
Can all irrational numbers be expressed by infinite number series? If so, can any transcendental equation have analytic solutions in the form of series?
$$e^{x}+\sin(x)-3=0$$
$e^x$ and $\sin x$ both have very simple expansions into infinite series (with rational coefficients). So a series for a solution of $e^x+\sin x-3=0$ can be obtained by the technique of reversion of series. See, for example, http://mathworld.wolfram.com/SeriesReversion.html or https://en.wikipedia.org/wiki/Lagrange_inversion_theorem
The expansion for $\pi$ should read $\pi/4=1-\frac{1}{3}+\frac{1}{5}-...+\frac{(-1)^n}{2n-1}+...$ and this (inverse series) expansion can be obtained from the series reversion of the expansion of $\tan(x)$.
Observe that $y = e^x+\sin(x)$ does not have an inverse series expansion about $y=0$ because $y=e^x$ does not (its inverse is $\log(y)$). But you can compute the inverse series about other points.
Alternatively, you can compute the inverse series of $u=e^x+\sin(x)-1$ and evaluate the resulting series in $u$ at $u=2$. To make sense, one has to consider the radius of convergence of the resulting inverse series.
Here is a general result for small $b$, $a>0$, and other unknown restrictions. It uses factorial power $u^{(v)}$:
$$e^x+b\sin(cx)=a\implies x=\ln\left(a+\sum_{n=1}^\infty\sum_{k=0}^n\frac{\left(-\frac{i b}2\right)^n}{n!}\binom nk (-1)^ka^{2ick-icn-n+1}(ic(2k-n))^{(n-1)}\right)$$
Test here by substituting
Many numbers can be a series, but what is the inner sum?
But not every real number will have a "nice" or a "regular" form of expansion.
– Mark Bennet Jan 31 '20 at 06:12