Infinite sum of a modified Bessel function of the first kind

Question

I am trying to get that the following sum of a modified Bessel function of the first kind reduces like such: $$\sum_{k=1}^\infty k I_k(z)=\dfrac{z}{2}(I_0(z)+I_1(z))$$ but I cannot find this equation anywhere in this Wolfram page. I do not think that I have the wrong expression but am not sure even that this sum should converge.

Answer 1

HINT :

Starting from the next relationship that one can found in all handbooks of special functions : $$I_{\nu+1}(z)=I_{\nu-1}(z)-\frac{2\nu}{z}I_{\nu}(z)$$ $$\nu I_{\nu}(z)=\frac{z}{2}\left(I_{\nu-1}(z)-I_{\nu+1}(z)\right)$$ With integer index $\nu=k$ : $$\sum_{k=1}^\infty k I_{k}(z)=\sum_{k=1}^\infty\frac{z}{2}\left(I_{k-1}(z)-I_{k+1}(z)\right)$$ $\sum_{k=1}^\infty k I_{k}(z)=\frac{z}{2}\left( (I_0-I_2)+ (I_1-I_3)+(I_2-I_4)+(I_3-I_5)+ ... \right)$

After simplification of couples terms which annihilate :

$$\sum_{k=1}^N k I_{k}(z)=\frac{z}{2}\left( I_0(z)+I_1(z)\right)-\frac{z}{2}\left( I_N(z)+I_{N+1}(z)\right)$$

In case of convergence of the series for $N\to\infty$ : $$\sum_{k=1}^\infty k I_{\nu}(z)=\frac{z}{2}\left( I_0(z)+I_1(z)\right)$$

This isn't always the case. In case of $z=x$ real large, the asymptotic is : $$\frac{x}{2}\left( I_N(x)+I_{N+1}(x)\right)\:\sim\: \sqrt{\frac{x}{2\pi}}e^x$$ which doesn't tends to $0$.

In case of $z=x$ real small $\quad \frac{x}{2}\left( I_N(x)+I_{N+1}(x)\right)\:\sim\: \frac{x^{N+1}}{2^{N+1}N!}\quad$ tends to $0$ when $N\to\infty$.