If we let $(a_n)$ and $(b_n)$ be sequences, and suppose there exists $N\in \Bbb N$ such that $a_n=b_n$ for all $n>N$. Suppose $\sum a_n$ converges and $\sum_{n=1}^{\infty}a_n=S$. I am looking to calculate the sum of $\sum_{n=1}^{\infty}b_n$
My solution:
Here is my logic but I am not sure if this is correct thinking or not, looking for some help.
$\sum b_n = S - C$ and $C$ goes to $0$ so $S-(a_1+...+a_N-b_1-b_2-....-b_N)$
I don't understand what you mean with the $C$ going to $0$, but the final result is fine.
Another way of going about it would be:
$$\sum_{i=1}^\infty b_i = \sum_{i=1}^N b_i + \sum_{i=N+1}^\infty a_i = \sum_{i=1}^N b_i + \left(\sum_{i=1}^\infty a_i - \sum_{i=1}^{N} a_i\right) = \\ S - \sum_{i=1}^{N} a_i + \sum_{i=1}^N b_i$$
The two sums are finite sums so they exist and $\sum b_i$ also converges.
Since $\sum_{n = 1}^\infty a_n = S$ and $b_n = a_n$ for any $n > N$ we have $$\sum_{n = N + 1}^\infty b_n = S - \sum_{n = 1}^N a_n$$ Hence $$\sum_{n = 1}^\infty b_n = S - \sum_{n = 1}^N a_n + \sum_{n = 1}^N b_n$$ So what you've done is basically right but $C$ definitely goes NOT to $0$ (it is also a bit unclear how you defined it).
