Does $\sum^{\infty}_{1}\frac{1}{k+1} = \sum^{\infty}_{2}\frac{1}{k}$ diverge because $\sum^{\infty}_{1}\frac{1}{k}$ diverges? If you manipulate the index of a series does it still converge/diverge?
For example:
Does $\sum^{\infty}_{1}\frac{1}{k+1} = \sum^{\infty}_{2}\frac{1}{k}$ diverge because $\sum^{\infty}_{1}\frac{1}{k}$ diverges?

$$\sum^{\infty}_{1}\frac{1}{k+1}=\frac 12 +\frac13 +\frac 14+\ldots$$
$$\sum^{\infty}_{1}\frac{1}{k}=1+\frac 12 +\frac13 +\frac 14+\ldots$$
Hence $$\sum^{\infty}_{1}\frac{1}{k+1}=\sum^{\infty}_{1}\frac{1}{k}-1$$
Since $$\sum^{\infty}_{1}\frac{1}{k}\to \infty \implies \sum^{\infty}_{1}\frac{1}{k}-\color{blue}1 \to \infty \implies \sum^{\infty}_{1}\frac{1}{k+1 }\to \infty$$
