# 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. – Lucozade Nov 14 '18 at 16:36

## 1 Answer

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.