Can you find an infinite series where…
- … the terms are decreasing and the series converges?
Since this question asks about infinite series which converge, I need to be sure that any series I suggest does converge. The series I know most about are arithmetic and geometric series. Infinite arithmetic series can’t converge except for \(0+0+0+\dotsb\), but I know conditions under which geometric series converge, so I’ll focus on geometric series for now.
I know that a geometric series will converge if the common ratio between the terms, \(r\), satisfies \(-1<r<1\), so one way to meet the required conditions is to find a convergent geometric series whose terms are decreasing (that is, where the terms satisfy \(u_{n+1}\leq u_n\) for all positive integers \(n\)).
The signs of the terms will alternate if the common ratio is negative, so I need to have a common ratio between \(0\) and \(1\). Is this enough to ensure that the terms are decreasing?
For example, I could choose the geometric series
\(1+\tfrac{1}{2}+\tfrac{1}{4}+ \tfrac{1}{8}+\tfrac{1}{16}+\dotsb\)
\(3+1+\tfrac{1}{3}+ \tfrac{1}{9}+\tfrac{1}{27}+\dotsb\)
\(256+128+64+32+16+\dotsb\)
Can you find an infinite series where…
- … the terms are decreasing and the series does not converge?
This time I want to be sure my series do not converge. Continuing from part (a), if I choose a geometric series with common ratio \(r\) where \(r\leq -1\) or \(r\geq 1\) the series will not converge. And I can ensure that the terms decrease if I choose \(r>1\) and the terms are negative. For example, the terms in the geometric series \(-1 -2 -4 -8 -\dotsb\) are decreasing and this series does not converge.
As I’m looking for a series that doesn’t converge, I could instead consider an arithmetic series with negative common difference. It doesn’t matter whether the first term is positive or negative, so for example, I could choose \(5+2-1-4 - \dotsb\)
So far I’ve considered arithmetic and geometric series, but I don’t have to restrict myself to these. For example, the terms in the series \(1+ \tfrac{1}{2}+ \tfrac{1}{3}+ \tfrac{1}{4}+ \tfrac{1}{5}+\dotsb\) are decreasing. This is known as the harmonic series. Do you think this series converges?
What about the series \(2+\tfrac{3}{2}+\tfrac{5}{4}+\tfrac{9}{8}+\tfrac{17}{16}\dotsb\)? Does this series meet the conditions for part (b)?
Can you find an infinite series where…
- … the terms are increasing and the series converges?
As I need a convergent series, I’ll look for a geometric series that meets the condition on increasing terms. Building on ideas from part (a), I’ll need the common ratio to be between \(0\) and \(1\), but then the terms will only increase if they are all negative. For example, I could choose the geometric series \(-1 -\tfrac{1}{3} - \tfrac{1}{9}-\tfrac{1}{27}- \dotsb\) Does this series meet the required conditions?
If the terms in a series get closer to zero as \(n\) increases, does this mean the series converges?
What if the series in (a) must converge to \(4\)?
For (a) I chose to look for geometric series with common ratio between \(0\) and \(1\) to ensure that the sum of the terms converged and the signs of the terms didn’t alternate. If the first term is positive then all the other terms will be, and since \(0<r<1\), these terms will decrease.
I’ll keep focusing on geometric series, but to meet the extra constraint that the sum of the terms must be \(4\), I need to find first term \(a\) and common ratio \(r\) so that \(\tfrac{a}{1-r}=4.\) I could choose \(r=\tfrac{1}{2}\), which would require \(a=2.\) (This has the same common ratio as the first series I gave for (a), but the first term is different). I could also choose \(a=1\), which would then require \(r=\tfrac{3}{4}.\) How much flexibility it there over choice of \(a\) and \(r\)?
If you found a geometric series for (a), how could you modify it to make a new infinite series that converges to \(4\)?
What if the terms of the series in (b) must all be positive?
If the terms in my series for (b) need to be positive but decreasing, I can’t have an arithmetic series.
The terms in a geometric series of positive numbers will only decrease if the common ratio is between \(0\) and \(1\), but then the sum of the terms will converge, so a geometric series cannot meet this extra condition in (b).
The harmonic series \(1+\tfrac{1}{2}+\tfrac{1}{3}+ \tfrac{1}{4}+\tfrac{1}{5}+\tfrac{1}{6}+\tfrac{1}{7}+\tfrac{1}{8}+\dotsb\) is decreasing and doesn’t converge, and all the numbers are positive, so this series already satisfies the extra constraint for (b). The series \(2+\tfrac{3}{2}+\tfrac{5}{4}+\tfrac{9}{8}+\tfrac{17}{16}\dotsb\) also meets these constraints.
If you add the same constant to each of the terms in the series you found in (c), will your new series still meet the conditions in part (c)?
For (c) I chose the geometric series \(-1 - \tfrac{1}{3}- \tfrac{1}{9}- \tfrac{1}{27}- \dotsb\), which consists of negative terms that increase to \(0\) because of the size of the common ratio. Adding \(1\) to every term gives the new series
\[0 + \tfrac{2}{3}+ \tfrac{8}{9}+ \tfrac{26}{27}+ \dotsb\]
Now instead of the terms increasing to \(0\) they increase to \(1\). But what can you say about the sum of the terms?
We have mainly focused on arithmetic and geometric series here because we already know quite a lot about how these behave. There are lots of other series that meet the conditions in this question but which require some intricate analysis to determine whether they converge.
In the final part of the problem we asked what would happen if you added a constant to every term in a series. How else could you modify geometric sequences or series and how does this affect their behaviour? For example, what happens if you multiply every term by the same constant? What if you square every term?
Here is some Student work for parts of this problem. How does this compare with your approach?
