Real analysis question about convergent series $\sum{a_k}$

Question

I would love any help with this question as I am pretty stuck.

Suppose that $\sum_{k=1}^{\infty}a_k$ is a convergent series.

Prove that the series $\sum_{k=n}^{\infty}a_k$ converges for each positive integer n.

Answer 1

Notice that $$\sum_{k=1}^{\infty}a_k-\sum_{k=1}^{n-1}a_k=\lim_{m\rightarrow \infty}(\sum_{k=1}^ma_k-\sum_{k=1}^{n-1}a_k)=\lim_{m\rightarrow\infty}\sum_{k=n}^ma_k=\sum_{k=n}^{\infty}a_k$$ which gives the result.

Answer 2

If $s:=\sum_{k=1}^\infty a_k$ is convergent, then by definition we have $s_n \to s$ where $s_n = \sum_{k=1}^n a_k$. This means (again by definition) that for all $\varepsilon >0$ we have a number $N\in \mathbf N$ such that $$|s-s_n| < \varepsilon \qquad \text{for all } n\geq N.$$ Since $s-s_n = \sum_{k=n+1}^\infty a_k$, we are done.

Answer 3

Here is another way to see it that may be slicker. View the series as a sequence of partial sums. It is well-known that such a sequence converges iff it is Cauchy. Since the original sum is convergent, it's terms are Cauchy. Thus, for any $\epsilon$, there is $N$ such that $m,n > N \implies \sum_n^m a_i < \epsilon$. This is clearly true even if we delete any number of terms at the beginning.