[proofplan] We compare the partial sums of the original series with the partial sums of the tail. The only difference between them is the fixed finite sum of the first $N$ terms. Since adding or subtracting a fixed scalar preserves convergence of a sequence in $\mathbb{R}$ or $\mathbb{C}$, convergence of one sequence of partial sums is equivalent to convergence of the other. [/proofplan] [step:Relate the original partial sums to the tail partial sums] Let $S_k \in \mathbb{K}$ denote the $k$-th partial sum of the original series, defined for $k \in \mathbb{N}$ by \begin{align*} S_k := \sum_{n=1}^{k} a_n. \end{align*} Let $T_k \in \mathbb{K}$ denote the $k$-th partial sum of the tail series, defined for $k \in \mathbb{N}$ by \begin{align*} T_k := \sum_{m=1}^{k} a_{N+m}. \end{align*} Finally, define the finite initial sum $A_N \in \mathbb{K}$ by \begin{align*} A_N := \sum_{n=1}^{N} a_n. \end{align*} For every $k \in \mathbb{N}$, splitting the finite sum at the index $N$ gives \begin{align*} S_{N+k} &= \sum_{n=1}^{N+k} a_n \\ &= \sum_{n=1}^{N} a_n + \sum_{n=N+1}^{N+k} a_n \\ &= A_N + \sum_{m=1}^{k} a_{N+m} \\ &= A_N + T_k. \end{align*} Therefore \begin{align*} T_k = S_{N+k} - A_N \end{align*} for every $k \in \mathbb{N}$. [/step] [step:Deduce convergence of the tail from convergence of the original series] Assume that $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{K}$. By definition, the sequence of partial sums $(S_k)_{k=1}^{\infty}$ converges to some $S \in \mathbb{K}$. Let $\varepsilon > 0$. Since $S_k \to S$, there exists $K_0 \in \mathbb{N}$ such that for every $j \geq K_0$, \begin{align*} |S_j - S| < \varepsilon. \end{align*} If $k \in \mathbb{N}$ satisfies $N+k \geq K_0$, then using the identity from the previous step gives \begin{align*} |T_k - (S - A_N)| &= |S_{N+k} - A_N - S + A_N| \\ &= |S_{N+k} - S| \\ &< \varepsilon. \end{align*} Hence $T_k \to S - A_N$ in $\mathbb{K}$. Therefore the tail series $\sum_{n=N+1}^{\infty} a_n$ converges. [/step] [step:Deduce convergence of the original series from convergence of the tail] Assume that the tail series $\sum_{n=N+1}^{\infty} a_n$ converges in $\mathbb{K}$. By definition, the sequence of tail partial sums $(T_k)_{k=1}^{\infty}$ converges to some $T \in \mathbb{K}$. Let $\varepsilon > 0$. Since $T_k \to T$, there exists $K_1 \in \mathbb{N}$ such that for every $k \geq K_1$, \begin{align*} |T_k - T| < \varepsilon. \end{align*} For every integer $j \geq N+K_1$, define $k := j - N$. Then $k \in \mathbb{N}$ and $k \geq K_1$. Using the identity $S_{N+k} = A_N + T_k$, we obtain \begin{align*} |S_j - (A_N + T)| &= |S_{N+k} - (A_N + T)| \\ &= |A_N + T_k - A_N - T| \\ &= |T_k - T| \\ &< \varepsilon. \end{align*} Thus $S_j \to A_N + T$ as $j \to \infty$. Therefore the original series $\sum_{n=1}^{\infty} a_n$ converges. [/step] [step:Conclude the equivalence] We have shown that convergence of $\sum_{n=1}^{\infty} a_n$ implies convergence of $\sum_{n=N+1}^{\infty} a_n$, and that convergence of $\sum_{n=N+1}^{\infty} a_n$ implies convergence of $\sum_{n=1}^{\infty} a_n$. Hence the two series converge simultaneously. [/step]