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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!

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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.

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    $\begingroup$ May I suggest that you avoid stating that a result is "obvious" or "trivial"? It can come across as rather condescending. If the OP had thought that the result was obvious, they probably wouldn't have asked the question. Telling them that it is obvious is essentially telling them that they are being dumb. Why muddy a perfectly adequate answer with implicit insults? $\endgroup$
    – Xander Henderson
    Jun 23 '18 at 16:23
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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 $n,m\to \infty$.

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    $\begingroup$ For OPs benefit, there's an extra trivial justification hidden here, in the second equality. The difference of the limits of each sum is the limit of the difference, which makes the sum finite when cancelling terms, precisely because the tail limit exists. $\endgroup$
    – Alex R.
    Jun 23 '18 at 5:35
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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.

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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.

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

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  • $\begingroup$ This is not really a proof. You're assuming the limit of the sum exists by declaring it to be $A$. $\endgroup$
    – Alex R.
    Jun 23 '18 at 5:31
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    $\begingroup$ The limit on the RhS exist by assumption. $\endgroup$ Jun 23 '18 at 5:36
  • $\begingroup$ No, the tail limit exists by assumption. The question is to prove the limit of the sum exists and is equal to some $A$. $\endgroup$
    – Alex R.
    Jun 23 '18 at 5:36

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