Let $i>0$ and $j>0$. Is it true that
$$\dfrac{\sum \limits_{k=i+j+1}^{\infty} p_k}{\sum \limits_{k=i+1}^{\infty} p_k} = \sum \limits_{k=j+1}^{\infty} p_k?$$
Let $i>0$ and $j>0$. Is it true that
$$\dfrac{\sum \limits_{k=i+j+1}^{\infty} p_k}{\sum \limits_{k=i+1}^{\infty} p_k} = \sum \limits_{k=j+1}^{\infty} p_k?$$
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As noted, this is false in general. It is correct when $p_k = \frac{1}{2^k}$, however.
$$ \sum_{k=j+1}^\infty \frac{1}{2^k} = \frac{1}{2^j} \\ \sum_{k=i+1}^\infty \frac{1}{2^k} = \frac{1}{2^i} \\ \sum_{k=i+j+1}^\infty \frac{1}{2^k} = \frac{1}{2^{i+j}} $$
More generally, it is true when $p_k = p^{k-1}(1-p)$ with $0 \le p \le 1$. Perhaps that was assumed in your source.