1

Although it's quite a trivial question but is it always that the very first term in a converging series is the largest term of all the terms in that series ?

Since if $\sum A_n$ converges, that is if $\sum A_n = a_0 + a_1 + a_2 +......$ converges then probably $a_n$ must be growing smaller and smaller as $n$ increases . If not then what are the exceptions ?

Arnav Das
  • 753
  • 3
    The important point of the two answers is that convergence is determined by what happens "far out to the high terms" rather than anything that happens low down. In this sense, low down can be quite high. Think of $a_n=\frac 1{2^n}$ except $a_{googol}=5$. This still converges. – Ross Millikan Oct 23 '15 at 10:35
  • 1
    Consider $0 + 1/2 + 1/2^2 + 1/2^3 + \cdots $ – zhw. Oct 23 '15 at 10:56
  • @zhw just a minor question : in the example you gave above, $0$ can't be procured from $1/{2^n}$ yet $0$ could be considered a part of that sequence ? – Arnav Das Oct 24 '15 at 09:53
  • Yes, the sequence is $0,1/2,1/2^2, \cdots$ – zhw. Oct 24 '15 at 16:40

4 Answers4

2

For a series $\sum a_n$ to converge, you need that $a_n\to 0$ as $n\to\infty$. But that does not mean you need to have $a_n>a_{n+1}$ for all $n$. For instance, the series $$\sum_{n=1}^{\infty}\frac{1}{n^2}\frac{\sin(n)}{n}$$ converges, but have a look at the first 20 values:

enter image description here

Edit: In fact, there is no need of an alternating series. For instance, $$\sum_{n=1}^{\infty}\frac{1}{n^2}\frac{k+\sin(n)}{n},\quad\mbox{with }k\ge 1.$$ Take a look at the case $k=1$.

enter image description here

AugSB
  • 5,007
  • True but its an alternating series, what would be the case if it were an converging series where all the terms would be positive ? – Arnav Das Oct 23 '15 at 10:07
  • It's the same. I edited my solution with another example. – AugSB Oct 23 '15 at 10:23
  • An unrelated query : how did you get the figure of graphs here ? – Arnav Das Oct 23 '15 at 16:48
  • @Arnas Mathematica – AugSB Oct 23 '15 at 16:50
  • OK what if the sequence were diverging ? Does it run akin to the answers above that largest term does exist but it need not be the last one ? – Arnav Das Oct 24 '15 at 09:57
  • @ArnavDas You can always take whatever your favourite (converging or diverging) series, change one of its terms (the first one, the 100th one... any) and the series will still maintain its condition. Maybe it will not have a general formula (I guess that is what you are looking for), but it will still be a series. – AugSB Oct 26 '15 at 09:48
  • yeah I meant that – Arnav Das Oct 26 '15 at 10:56
2

A very simple counterexample is the series $$0+0+…+1+0+0+…$$ That is, for each $n$, we have the sum $1 = \sum_{k=1}^∞ a^{(n)}_k$ of the sequence $a^{(n)}_k = \begin{cases}1 & k=n \\ 0 & k≠n\end{cases}$, which can be made to have the $k_0$th term $a^{(n)}_{k_0}$ biggest by choosing $n=k_0$.

Calvin Khor
  • 34,903
1

All you can say is that the largest term occurs in a finite place, not necessarily the first. Think of a convergent series, now add an extra term, larger than all terms in the original sequence and make it your new $a_{100}$, the new series will converge to the original limit plus this new term, but the largest term will be in the $100^{th}$ place.

Ross Millikan
  • 374,822
Shahar Even-Dar Mandel
  • 1,158
  • Can you give an example ? and what do you exactly mean by a finite place ? – Arnav Das Oct 23 '15 at 09:59
  • I edited my answer – Shahar Even-Dar Mandel Oct 23 '15 at 10:03
  • Well think if I considered the series $a_n = {\sin x}/x$ how come I add a term which is greater then all the terms in this series since even if I do it would not be a part of the $a_n$. Another question(not related to the above question)if also somehow if I did add a new term to the series the limit of the series would still be converging to same original limit why would it tend to some another term as you say that is 'converge to the original limit plus this new term' ?? – Arnav Das Oct 23 '15 at 10:26
  • A $\LaTeX$ hint: for multidigit things like subscripts, you need to enclose them in braces. To get $a_{100}$ you put a_{100} between dollar signs. – Ross Millikan Oct 23 '15 at 10:32
1

The exponential series $$\exp(z):=\sum_{k=0}^\infty{1\over k!}\>z^k\qquad(z\in {\mathbb C})$$ provides an infinity of examples. It converges for any given $z\in{\mathbb C}$, but the position of its largest term (in absolute value) depends heavily on the chosen $z$: Put $$a_k:=\left|{z^k\over k!}\right|\ .$$ Then $${a_k\over a_{k-1}}={|z|\over k}$$ is $>1$ as long as $k<|z|$, and $<1$ when $k>|z|$. Therefore the $a_k$ are increasing at the start, reach a maximum when $k$ is is one of $\bigl\lfloor |z|\bigr\rfloor$ and $\bigl\lceil |z|\bigr\rceil$, and then decrease monotonically to $0$.

Christian Blatter
  • 226,825