series including infinite sum I am looking for the approximation of the following function:
$$\rho(a,b)=1-e^{-(a+b)}\sum_{m=1}^{\infty}\left(\sqrt{\frac{b}{a}}\right)^m I_m(2\sqrt{ab})$$
where $I_m(x)$ is the modified Bessel function. Since $I_m(x)$ is again an infinite series, up to how many terms I need to do summation in both the cases above in order to get some better approximation? Is there any rule of thumb?
 A: If one is interested in expressing $\rho(a,b)$ with a single summation/integration operation, one can use, for every positive $a$ and $b$ such that $b<a$,
$$
\rho(a,b)=1+e^{-(a+b)}\int\limits_0^\pi e^{2\sqrt{ab}\cos\theta}\frac{b-\sqrt{ab}\cos\theta}{a+b-2\sqrt{ab}\cos\theta}\frac{\mathrm{d}\theta}\pi.
$$
See below for some explicit formulas when $a=b$ or $a<b$.
Case $b< a:$ This uses only the integral formula for modified Bessel functions with integer parameter, which says that for every integer $m$ and every $x$ with nonnegative real part,
$$
I_m(x)=\int\limits_0^\pi e^{x\cos\theta}\cos(m\theta)\frac{\mathrm{d}\theta}\pi.
$$
Writing the cosine as the real part of a complex exponential and summing this over $m\ge1$ yields, for every real number $s$ such that $|s|<1$,
$$
\sum\limits_{m\ge1}s^mI_m(x)=\text{Re}\int\limits_0^\pi e^{x\cos\theta}\frac{se^{i\theta}}{1-se^{i\theta}}\frac{\mathrm{d}\theta}\pi.
$$
Multiplying the numerator and the denumerator by the conjugate quantity $1-se^{-i\theta}$ and using the relation $|1-se^{i\theta}|^2=1+s^2-2s\cos\theta$, one sees that
$$
\sum\limits_{m\ge1}s^mI_m(x)=\int\limits_0^\pi e^{x\cos\theta}\frac{s\cos\theta-s^2}{1+s^2-2s\cos\theta}\frac{\mathrm{d}\theta}\pi.
$$
Plugging in $x=2\sqrt{ab}$ and $s=\sqrt{b/a}$ (hence the restriction $b< a$) yields $\rho(a,b)$ as above.
Case $a=b$: One sees that
$$
\rho(a,a)=1-e^{-2a}\sum\limits_{m\ge1}I_m(2a).
$$
Using the fact that $I_m(x)=I_{-m}(x)$ for every integer $m$ and the formula (written somewhere in the WP page already mentioned) for the generating function
$$
\sum\limits_{m=-\infty}^{+\infty}I_m(x)\cos(m\theta)=e^{x\cos\theta},
$$
at $\theta=0$, one sees that 
$$
\rho(a,a)=\frac12+\frac12e^{-2a}I_0(2a)=\frac12+\frac12e^{-2a}\int\limits_0^\pi e^{2a\cos\theta}\frac{\mathrm{d}\theta}\pi.
$$
Case $a<b$: Comparing $\rho(a,b)+\rho(b,a)$ with the generating function of the family $(I_m(x))$ that we recalled above, one gets
$$
\rho(a,b)+\rho(b,a)=1+e^{-(a+b)}I_0(2\sqrt{ab}),
$$
for every $a$ and $b$. Hence, when $a< b$,
$$
\rho(a,b)=e^{-(a+b)}\int\limits_0^\pi e^{2\sqrt{ab}\cos\theta}\frac{b-\sqrt{ab}\cos\theta}{a+b-2\sqrt{ab}\cos\theta}\frac{\mathrm{d}\theta}\pi.
$$
Remark: The result is discontinuous around $a=b$ since, for every fixed $a$, $\rho(a,b)\to\rho(a,a)+\frac12$ when $b\to a^-$ and $\rho(a,b)\to\rho(a,a)-\frac12$ when $b\to a^+$.
