If the sum of the tail of a series goes to $0$, must the series converge?

Question

Suppose $\{a_n\}$ is a sequence of positive terms.

It is a well known result that if the series $\displaystyle \sum_{k=1}^{\infty} a_k$ converges then $\displaystyle \lim_{m \rightarrow \infty} \displaystyle \sum_{k=m}^{\infty} a_k=0 $.

Is the converse true?

That is, is it true that if the tail of a series goes to zero, then the series must converge?

My thoughts:

If we let $S_n=\displaystyle \sum_{k=1}^{n} a_k$, then clearly $S_n$ is an increasing sequence, and for $n > m$ , we have $S_n - S_m = \displaystyle \sum_{k=m+1}^{n} a_k $

Then we want to show that if $\displaystyle \lim_{m\rightarrow \infty} (\displaystyle \lim_{n\rightarrow \infty} S_n - S_m ) = 0$ (or just if it exists) then $\displaystyle \lim_{n\rightarrow \infty} S_n < \infty$.

Note that since $S_n$ is increasing, it is enough to show that it is bounded.

I tried to show this by definition, but my issue is that I am not sure how to deal with that double limit.

Any help would be really appreciate it.

Thanks!

Answer 1

Yes, it is correct also when the sequence $(a_n)_n$ is not positive because $(S_n)_n$ is a Cauchy sequence $$|S_n - S_m| = | \sum_{k=m+1}^{n} a_k|=|\sum_{k=m+1}^{\infty} a_k-\sum_{k=n+1}^{\infty} a_k|\to 0$$ as $n,m\to \infty$.

Answer 2

The result you want is obvious. If $$\lim_{m\to\infty } \sum_{n=m}^{\infty} a_n = 0$$ then for some $M$ we have that $$\sum_{n=M}^{\infty} a_n <\infty.$$ Tacking on the first $M$ terms doesn't affect convergence.

Answer 3

The answer is yes.

Since the tail approaches $0$, we can find an $N$ such that $$ -1<\sum_{k=N+1}^{\infty} a_k<1$$

Note that $$\displaystyle \sum_{k=1}^{\infty} a_k=\displaystyle \sum_{k=1}^{N} a_k + \displaystyle \sum_{k=N+1}^{\infty} a_k =A+\displaystyle \sum_{k=N+1}^{\infty} a_k < \infty $$

Answer 4

For the statement "$\displaystyle \lim_{m \rightarrow \infty} \displaystyle \sum_{k=m}^{\infty} a_k=0$" to make any sort of sense we have to have that for some $M$, $\displaystyle \sum_{k=m}^{\infty} a_k$ exists for all $m\geqslant M$; we just can't talk about the limit otherwise.

Now let $(b_n)$ be the (tail) sequence $(a_{n+M})$.

Let $S_n=\sum_{k=0}^{n} a_k$, and $T_n=\sum_{k=0}^{n} b_k$. We know $\displaystyle \sum_{k=M}^{\infty} a_k=\displaystyle \sum_{k=0}^{\infty} b_k$ exists, which means by definition that $\lim_{n\to\infty}T_n$ exists.

Then for $n \geqslant M$ we have $S_n=(a_0+\dots+a_M) +T_{n-M}$. As the limit of the right hand side exists, so too does the limit of the left hand side and we are done.

Answer 5

Here is D. Brogan's and Mohammad Riazi-Kermani's proof written out with all the details I think are necessary.

Since $\displaystyle\lim_{m\to\infty}\sum_{n=m}^\infty a_n=0$, there is an $M$ such that $\displaystyle\sum_{n=M}^\infty a_n$ is finite, say it's equal to $\tilde L$. For $k\geq M$ let $\displaystyle\tilde S_k=\sum_{n=M}^k a_n$. Then we have, for $k\geq M$, $\displaystyle S_k=\sum_{n=1}^{M-1}a_n+\tilde S_k$. The right hand side is a sum of a constant and a convergent sequence, hence $S_k$ is also a convergent sequence, i.e. the original sum converges.