Linear non-homogeneous recurrence solution?

Question

I have the following equation:

$$A_{n}=A_{n-1}+\frac{1}{\sqrt{2n-1}}$$

with the inital conditions $A_1=1$ and $A_2= 1 +\frac{1}{\sqrt{3}}$

Does anyone know how to find the closed form expression?

Answer 1

Probably too complex.

The general solution of $$A_{n}=A_{n-1}+\frac{1}{\sqrt{2n-1}}$$ is $$A_n=c_1-\frac{\zeta \left(\frac{1}{2},n+\frac{1}{2}\right)}{\sqrt{2}}+\left(1-\frac{1}{\sqrt{2}} \right) \zeta \left(\frac{1}{2}\right)$$ where appear the Hurwitz zeta function and the zeta function.

Using $A_1=1$, we then have $$A_n=1+\frac{\zeta \left(\frac{1}{2},\frac{3}{2}\right)}{\sqrt{2}}-\frac{\zeta \left(\frac{1}{2},n+\frac{1}{2}\right)}{\sqrt{2}}=1+\frac{\zeta \left(\frac{1}{2},\frac{3}{2}\right)-\zeta \left(\frac{1}{2}\right)}{\sqrt{2}}+\frac{1}{\sqrt{2}}H_{n-\frac{1}{2}}^{\left(\frac{1}{2}\right)}$$ where appear generalized harmonic numbers.

Using, for large values of $p$, the asymptotics of harmonic numbers $$H_{p}^{\left(\frac{1}{2}\right)}=\zeta \left(\frac{1}{2}\right)+2 \sqrt{p}+\frac{1}{2 \sqrt{p}}+O \left(\frac{1}{p^{3/2}}\right)$$ we end with $$A_n=1+\frac{\zeta \left(\frac{1}{2},\frac{3}{2}\right)}{\sqrt{2}}+\sqrt{2n-1}+\frac 1{ 2\sqrt{2n-1}}+O\left(\frac{1}{n^{3/2}}\right)$$

Let $n=10^k$ and compute to get $$\left( \begin{array}{ccc} k & A_{10 ^k} \approx & A_{10 ^k} =\\ 1 & 4.045878878 & 4.044873362 \\ 2 & 13.71445211 & 13.71442242 \\ 3 & 44.29363302 & 44.29363208 \\ 4 & 140.9936283 & 140.9936283 \end{array} \right)$$