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I remember a teacher telling me that the absolute value of any term in a sequence will exceed some forms of sum of the infinitely many terms after it. I do not remember if it was 'the absolute value of the sum of the remaining terms', which I recognise as being the "looser" case, or if it was 'the sum of the absolute value of each of the remaining terms'. I doubt this is the case as I know not all convergent series are absolutely convergent.

Although actually, I don't quite remember if it was that the absolute value of the sum of the first nth terms exceeds the absolute value of the sum of the remaining terms. This is a possibility.

I recognise that for an alternating series, the sum of all terms is bounded by the value of the first term, although this is only for alternating series.

I apologise for this being so vague. I just want to learn more about series and I think knowing such properties will greatly help with problems. I'm I could reason why these properties arise if I knew them, but coming up with them myself when faced with a problem is a completely different matter! I believe that this property that I am referring to in the question was used understanding Taylors theorem for the error of truncation of an infinite series, although I believe this can also be considered in terms of the mean value theorem.

EDIT

For example, if you have an infinite series

$S =a_1 +a_2 + a_3 +...$

Then perhaps $|\Sigma _{i=3} ^{\infty}a_i| < |a_2|$?

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    Can you give an example or rephrase this? I'm not sure what you're asking. Also, you can have a convergent alternating series where the terms aren't all decreasing, you only need them to be *eventually* going to zero. – 2017-01-11
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    That can't possibly be true, because I can take any convergent series and replace $a_2$ with zero. The series remains convergent but it violates your property. – 2017-01-11

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I think you are rembering the case where the alternating series test applies: If $a_n$ is a decreasing nonnegative sequence, with $a_n\to 0,$ then $\sum (-1)^na_n$ converges. Moreover,

$$a_n \ge \left|\sum_{k=n+1}^{\infty} (-1)^ka_k \,\right|$$

for all $n.$

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    Ah I see. The modulus of the nth term of an alternating series is greater than or equal to the modulus of sum of the remaining terms provided the series converges. Thank you for clarifying. – 2017-01-11
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Take $a_n=\frac{1}{n^2}$

we have $a_2=\frac{1}{4}$ but

$\sum_{n=3}^{+\infty}a_n=\frac{\pi^2}{6}-1-a_2>a_2$.

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Recall the famed geometric series to see that

$$\frac1{2^0}=\sum_{n=1}^\infty\frac1{2^n}$$

Other values will quickly show your statement is not true.

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Your assertion doesn't hold for all infinite series. However, a modification of your assertion is valid in special cases. For example, if you can find a positive constant $r<1$ such that $$ \left|\frac{a_{n+1}}{a_n}\right|\le r $$ for all large $n$, then we have $$ |a_{n+k}|\le r^k|a_n|\tag1 $$ for all large $n$ and therefore $$ \begin{align} |a_{n+1} + a_{n+2} + a_{n+3}\cdots|&\le |a_{n+1}| + |a_{n+2}| + |a_{n+3}|\cdots\\ &\stackrel{(1)}\le r|a_n| + r^2|a_n| + r^3|a_n| + \cdots\\ &=(r+r^2+r^3+\cdots)|a_n|\\ &=\frac r{1-r}|a_n|. \end{align} $$ It follows that the series $\sum a_i$ converges absolutely. (This argument is used in the proof of the ratio test.)