Does each term in an infinite series have to be greater than the infinite sum after it?

Question

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

Answer 1

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 $n.$

Answer 2

Take $a_n=\frac{1}{n^2}$

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

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

Answer 3

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.

Answer 4

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.)