Comprehensible lower bound for $1 - \sum_{n=0}^{d-1} \frac{\lambda^{nd}}{1+(n+1)(d-1)}$ with $\lambda \in (0,1)$

Question

I have the following expression for $\lambda \in (0,1)$ and $d \in \{2,3,\dots\}$: $$A(\lambda,d) := 1 - \sum_{n=0}^{d-1} \frac{\lambda^{nd}}{1+(n+1)(d-1)},$$ clearly we always have $A(\lambda,d) > 0$, and as $\lambda \rightarrow 1$, we have $A(\lambda,d) \rightarrow 0$ and as $\lambda \rightarrow 0$, we have $A(\lambda,d) \rightarrow 1 - \frac{1}{d}$. However, I wonder what happens in between.

Answer 1

take $g(x)=\sum_{n=0}^{d-1}{{\lambda^{nd}}\over{1+(n+1)(d-1)}}=\sum_{n=0}^{d-1}{{\lambda^{nd}}\over{nd+d-n}}$ therefore: $${1\over{\lambda^d}}{\sum_{n=0}^{d-1}{{\lambda^{(n+1)d}}\over{(n+1)d}}}\le g(x)\le {1\over d}+{\sum_{n=1}^{d-1}{{\lambda^{nd}}\over{nd}}}$$ Now define $f_1(\lambda;d)={\sum_{n=1}^{d}{{\lambda^{nd}}\over{nd}}}$ and $f_2(\lambda;d)={1\over d}+{\sum_{n=1}^{d-1}{{\lambda^{nd}}\over{nd}}}$ by differentiating we get: $${{df_1(\lambda;d)}\over{d(\lambda)^d}}={1\over {d}}\sum_{n=1}^{d}{(\lambda^d)^{n-1}}={1\over d}{{{1-\lambda^{d^2}}}\over{1-\lambda^d}}$$ $${{df_2(\lambda;d)}\over{d(\lambda)^d}}={1\over d}{{{1-\lambda^{d(d-1)}}}\over{1-\lambda^d}}$$ Then we have: $$f_1(\lambda;d)={1\over {d}}\int_{0}^{\lambda^d}{{1-u^d}\over{1-u}}du$$ $$f_2(\lambda;d)={1\over d}+{1\over d}\int_{0}^{\lambda^d}{{1-u^{d-1}}\over{1-u}}du$$ The functions $f_1(x;d)$ and $f_2(x;d)$ have no closed forms then they can be bounded by

$$f_1(x;d)>{{\lambda^{d^2}}\over{d^2}}$$ and $$f_2(x;d)<{1\over d}\ln {{e}\over{1-\lambda^d}}$$

so we can write

$$1-{1\over d}\ln {{e}\over{1-\lambda^d}}<A(x,d)<1-{{\lambda^{d^2}}\over{d^2}}$$