Determining whether $\sum_{i=1}^\infty \frac{t^{i}}{{i}!} (\sum_{j=0}^{i-1} k^j)$ converges

Let's consider a map $$L$$ between complex matrices: $$L(A) = kA + B$$ for some real value $$k$$ and matrix $$B$$. I am trying to calculate $$e^{tL}(A)$$ for some real value $$t>0$$. Here is my attempt:

Since $$L^n(A) = k^nA + \big(\sum_{i=0}^{n-1}k^i\big)B$$, we have:

$$e^{tL}(A) = A + tL(A) + \frac{t^2}{2!}L^2(A) + \frac{t^3}{3!}L^3(A) + \dotsc$$

$$= A + t(kA + B) + \frac{t^2}{2!}\big(k^2A + (k+1)B\big) + \frac{t^3}{3!}\big(k^3A + (k^2+k+1)B\big) + \dotsc$$

$$= \big(1 + k + \frac{(kt)^2}{2!} + \dotsc\big)A + \big(t + \frac{t^2}{2!}(k+1) + \frac{t^3}{3!}(k^2+k+1) + \dotsc \big)B$$

$$= e^{kt}A + \big(t + \frac{t^2}{2!}(k+1) + \frac{t^3}{3!}(k^2+k+1) + \dotsc \big)B$$.

I am not sure how to calculate the coefficient for the matrix $$B$$, which is $$\sum_{i=1}^\infty \frac{t^{i}}{{i}!} (\sum_{j=0}^{i-1} k^j)$$.

Does it converge/diverge? If it diverges, is it possible to obtain a condition on $$k$$ such that it converges?

We have that by geometric series with $$k\neq 1$$

$$\sum_{j=0}^{i-1} k^j=\frac{k^i-1}{k-1}$$

and then

$$\sum_{i=1}^\infty \frac{t^{i}}{{i}!} \left(\sum_{j=0}^{i-1} k^j\right)=\sum_{i=1}^\infty \frac{t^{i}}{{i}!} \frac{k^i-1}{k-1}=\frac{1}{k-1}\sum_{i=1}^\infty \frac{(tk)^{i}}{{i}!}-\frac{1}{k-1}\sum_{i=1}^\infty \frac{t^{i}}{{i}!}=$$

$$=\frac{e^{tk}-1}{k-1}-\frac{e^t-1}{k-1}=\boxed {\frac{e^{tk}-e^t}{k-1}}$$

For the case $$k=1$$

$$\sum_{j=0}^{i-1} k^j=\sum_{j=0}^{i-1} 1=i$$

and then

$$\sum_{i=1}^\infty \frac{t^{i}}{{i}!} \left(\sum_{j=0}^{i-1} k^j\right)=\sum_{i=1}^\infty \frac{t^{i}}{{i-1}!}=t\sum_{i=1}^\infty \frac{t^{i-1}}{{i-1}!}=\boxed {te^t}$$

• And for $k=1$ it is $te^t$.
– user
Jul 3, 2023 at 20:32
• @user Thanks, I add also a derivation for that!
– user
Jul 3, 2023 at 20:57