# Convergence and divergence of series when changing finite number of summands

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.

• 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. Nov 14 '18 at 16:36

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.