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I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:

Let $\sum_{k=n_1}^{\infty} a_k$ and $\sum_{k=n_2}^{\infty}b_k$ be two series with $(s_n)_{n \geq n_1}$ and $(t_n)_{t \geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $k\geq N$, than we have

\begin{align}s_n = \sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + \ldots + a_{N-1} + \sum_{k=N}^{n}a_k\end{align} and

\begin{align}t_n &= \sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + \ldots + b_{N-1} + \sum_{k=N}^{n}a_k \\ &= s_n - \left(a_{n_1} + a_{n_1+1} + \ldots + a_{N-1}\right) + \left(b_{n_2} + b_{n_2+1} + \ldots + b_{N-1}\right)\end{align}

for all $n \geq N$. Hence, both $(s_n)_{n \geq n_1}$ and $(t_n)_ {t\geq n_2}$ are either convergent or divergent.

Unfortunately I do not see why $(s_n)_{n \geq n_1}$ and $(t_n)_{t \geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series. Can someone please help me understanding this proof.

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    $\begingroup$ Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not. $\endgroup$ – Lucozade Nov 14 '18 at 16:36
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Suppose $\lim_{n \to \infty}t_n\to t$, and \begin{align}t_n &= s_n - \left(a_{n_1} + a_{n_1+1} + \ldots + a_{N-1}\right) + \left(b_{n_2} + b_{n_2+1} + \ldots + b_{N-1}\right).\end{align}

We can let $\left(a_{n_1} + a_{n_1+1} + \ldots + a_{N-1}\right) - \left(b_{n_2} + b_{n_2+1} + \ldots + b_{N-1}\right)=C,$ a constant that is independent of $n$, then we have

$$t_n = s_n - C$$

then we have

\begin{align}\lim_{n \to \infty}t_n &= \lim_{n \to \infty}s_n - C\end{align}

Hence \begin{align} \lim_{n \to \infty}s_n =t+ C\end{align}

That is $s_n$ converges as well.

Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.

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