Discovered the following relation: $$\sum _{k=1}^{\infty } \sigma (k) \left(K_2\left(4 \pi \sqrt{k+y} \sqrt{y}\right)-K_0\left(4 \pi \sqrt{k+y} \sqrt{y}\right)\right)=\frac{\pi K_1(4 \pi y)-3 K_0(4 \pi y)}{48 \pi ^2 y}$$
where $\sigma(x)$ is a divisors sum, $K_n(x)$ is Modified Bessel function of second kind and $y>0$.
Also each of sums $$\sum _{k=1}^{\infty } \sigma (k) K_2\left(4 \pi \sqrt{k+y} \sqrt{y}\right)$$ $$\sum _{k=1}^{\infty } \sigma (k) K_0\left(4 \pi \sqrt{k+y} \sqrt{y}\right)$$ converge.
The question here is: as the difference (the first sum) has a closed form, may we expect each of sums above to have the closed form either, and if so does someone know what it can be?
In order to partially satisfy @Greg Martin request let me just provide numerical examples which will show the correctness of first relation. As I don't know any simple way of proving it.
For $y=1$ $$\sum _{k=1}^{300} \sigma (k) \left(K_2\left(4 \pi \sqrt{k+1} \sqrt{1}\right)-K_0\left(4 \pi \sqrt{k+1} \sqrt{1}\right)\right) = 6.8126525731073335538349556626598125635572137980741872738568342124.10^{-10}$$
$$\frac{\pi K_1(4 \pi)-3 K_0(4 \pi)}{48 \pi ^2} = 6.8126525731073335538349556626598125635572137980741872738568342124.10^{-10}$$
For $y=0.2$ $$\sum _{k=1}^{300} \sigma (k) \left(K_2\left(4 \pi \sqrt{k+0.2} \sqrt{0.2}\right)-K_0\left(4 \pi \sqrt{k+0.2} \sqrt{0.2}\right)\right) = 0.000466622676964315708526403991580236142341156773881429672591795$$
$$\frac{\pi K_1(4 \pi 0.2)-3 K_0(4 \pi 0.2)}{48 \pi ^2 0.2} = 0.000466622676964315708526403991580236142344783982506129849211143$$
There is a promissive relation to Ramanujan's work.
