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

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 $n$.

My attempt thus far:

$$\sum_{k=1}^{\infty}a_k - \sum_{k=1}^{n-1}a_k = \lim_{m\to \infty} \left(\sum_{k=1}^{m}a_k - \sum_{k=1}^{n-1}a_k\right)= \lim_{n\to \infty} \left(\sum_{k=n}^{m}a_k\right) = \sum_{k=n}^{\infty}a_k$$

But I am not sure how to conclude from here out that it converges. Basically, I don't understand how to wind up this answer.

## 4 Answers

Here is a more rigorous proof.

You want to show that, for any $n > 0$, $\lim_{m \to \infty} \sum_{k=n}^{m} a_k$ exists.

This means that, for each $n$, there is a $L(n)$ such that $\lim_{m \to \infty} \sum_{k=n}^{m} a_k =L(n)$.

You aleady know that $\lim_{m \to \infty} \sum_{k=1}^{m} a_k$ exists. Call this $L$. Then, for any $n$ and $m$, $\sum_{k=n}^{m} a_k =\sum_{k=1}^{m} a_k-\sum_{k=1}^{n-1} a_k$. Therefore, if we define $L(n) =L-\sum_{k=1}^{n-1} a_k$, then $\sum_{k=n}^{m} a_k-L(n) =\sum_{k=n}^{m} a_k-(L-\sum_{k=1}^{n-1} a_k) =\sum_{k=1}^{m} a_k-L$ and we know that this latter difference goes to zero as $m \to \infty$.

Therefore $\lim_{m \to \infty}\sum_{k=n}^{m} a_k-L(n) = 0$.

You've noticed that if $m\geq n$ then $$\sum_{k=n}^{m}a_k = \sum_{k=1}^{m}a_k - \sum_{k=1}^{n-1}a_k\tag{1}$$ Recall that we say that the series $\displaystyle\sum_{k=n}^{\infty}a_k$ converges if its sequence of partial sums $\displaystyle\sum_{k=n}^{m}a_k$ (for $m\geq n$) converges. That is to say, if $\displaystyle\lim_{m\to\infty}\displaystyle\sum_{k=n}^{m}a_k$ exists (in $\mathbb{R}$). Now, in view of $(1)$, we have $$\lim_{m\to\infty}\sum_{k=n}^{m}a_k = \lim_{m\to\infty}\left(\sum_{k=1}^{m}a_k - \sum_{k=1}^{n-1}a_k\right)\tag{2}$$ By hypothesis $\displaystyle\lim_{m\to\infty}\sum_{k=1}^{m}a_k$ exists (hence is a real number). The term $\displaystyle\sum_{k=1}^{n-1}a_k$ is a constant (with respect to $m$) finite sum and is also a real number. Finally, we know that, for sequences, "the limit of the sum is equal to the sum of the limits". In particular, when the two separate limits exist (which is the case here), then the limit of the sum exists. Hence, in view of $(2)$, we conclude that the limit $\displaystyle\lim_{m\to\infty}\sum_{k=n}^{m}a_k$ exists. Moreover, you even found an expression for its value in terms of the value of $\displaystyle\lim_{m\to\infty}\sum_{k=1}^{m}a_k$: \begin{align*} \lim_{m\to\infty}\sum_{k=n}^{m}a_k &= \lim_{m\to\infty}\sum_{k=1}^{m}a_k - \lim_{m\to\infty} \sum_{k=1}^{n-1}a_k \\ &=\lim_{m\to\infty}\sum_{k=1}^{m}a_k - \sum_{k=1}^{n-1}a_k \tag{3} \end{align*} Note that by definition, using the usual notations, $(3)$ merely says that $$\sum_{k=n}^{\infty}a_k = \sum_{k=1}^{\infty}a_k - \sum_{k=1}^{n-1}a_k$$

Note: Your work is correct (except one of your $\displaystyle\lim_{n\to\infty}$ should be $\displaystyle\lim_{m\to\infty}$). You basically started from the end (unless you read from right to left!), but at each stage of your argument you had a valid equality and hence you can conclude that $\displaystyle\sum_{k=n}^{\infty}a_k$ exists and equals $\displaystyle\sum_{k=1}^{\infty}a_k - \sum_{k=1}^{n-1}a_k$. You implicitly used the hypothesis that $\displaystyle\sum_{k=1}^{\infty}a_k$ is convergent when you wrote $\displaystyle\lim_{m\to\infty}\sum_{k=1}^{m}a_k$.

Remember, that in your $\sum_{k=1}^{\infty}a_k$- $\sum_{k=1}^{\ n-1}a_k$

your first one already converges and left one is just a finite sum, which converges too. So there you go.

The really important concept is that convergence or divergence only depends on what happens "near infinity". Anything that impacts finitely many terms cannot change whether the sum converges or diverges. If you want to do an $\epsilon - N$ proof, you can use the same $N$ for all of your sums. Deleting a finite number of terms doesn't change whether the infinite sum converges, it just changes the sum by the sum of the deleted terms.