If $\sum_{k=0}^\infty a_k$ is convergent with value $s$, what about $\sum_{k=0}^\infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?
My reasoning: $$\sum_{k=0}^\infty b_k =\lim_{n \rightarrow \infty} \sum_{k=0}^n b_k=\lim_{n \rightarrow \infty} \sum_{k=0}^na_{k+1}-2a_{k+3} $$ Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards: $$ \lim_{n \rightarrow \infty} \sum_{k=0}^n b_k=\lim_{n \rightarrow \infty} \left( \sum_{k=0}^n a_{k+1}- \sum_{k=0}^n 2a_{k+3} \right) $$ Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet. $$\lim_{n \rightarrow \infty} \left( \sum_{k=0}^n a_{k} - a_0- 2\sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2\right) $$ Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get: $$\sum_{k=0}^\infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$
Did that all make sense, is my reasoning correct? conclusion: it converges as we computed the exact value.
I will write this in the answer section so that the question does not remain open.
The reasoning looks fine and correct to me.