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With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

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With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^n\binom{n}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

Hint: This answer might be useful.

With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

Hint: You might also find this answer useful.

With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

Hint: This answer might be useful.

With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

Hint: You might also find this answer useful.