# Does the nth term of the series give an over- or under-estimate for the n-tails?

I am a TA, and tomorrow, I have to give my students the following problem to work on:

Consider a convergent series $$s = \sum_{k=0}^{\infty} a_{k}.$$ By definition, $s$ is the limit as $n \to \infty$ of the partial sums $s_{n}=\sum_{k=0}^{n}a_{k} = a_{0}+a_{1}+\cdots + a_{n}$. As you calculate $s_{n}$ for bigger and bigger $n$, the first digits of $s_{n}$ will start matching up with $s$. The error $E_{n} = |s-s_{n}|$ gives how many digits. For example, if $E_{n} = 0.00035$ then the $s$ and $s_{n}$ will have the same first two decimals (unless $s$ is something like $0.999978$...).

Consider the $n+1$-tail of a series: $$s = s_{n}+t_{n+1} = \sum_{k=0}^{n}a_{k} + \sum_{k=n+1}^{\infty}a_{k} = s_{n} + t_{n+1}.$$ The tail $t_{n+1}$ precisely gives the error between the $n$th partial sum and the actual value $s$ of the series: $$E_{n} = s - s_{n} = t_{n+1}$$

Consider the convergent series with positive terms $\sum_{k=1}^{\infty} \frac{1}{k^{2}} = \sum_{k=1}^{\infty}a_{k}$. Does the $n$th term $a_{n}$ give an over- or under-estimate of the $n$-tails?

The solution we were provided with says that $a_{n}$ is less than the tail and therefore is an underestimate, but it does not include any details as to why this is the case. I have emailed my supervisor and asked him for clarification, but I am not sure I will hear from him anytime soon, and even if I do, whether his clarification will make things make any more sense.

I know that the series $\sum_{k=1}^{\infty} \frac{1}{k^{2}} = \frac{\pi^{2}}{6} \approx 1.644934$, but nowhere on the assignment does it mention this, nor am I aware of whether or not the students know this fact. I was attempting to find several partial sums, and I found $s_{1} = 1$, $s_{2} = \frac{5}{4} \approx 1.25$, $s_{3} = \frac{49}{36} \approx 1.36111$, $s_{4} = \frac{870}{576}\approx 1.4236$. But, in each of those $i$ cases, the tail would equal $\frac{\pi^{2}}{6} - s_{i}$, which is going to be less than $s_{i}$, so how are these under-estimates?

Unless I am approaching the problem completely wrong, which is a possibility.

Could somebody please help me understand how to do this problem, so that I can help a bunch of people who are far more clueless than I am figure out how to do it tomorrow? I thank you in advance for your time and patience!

• What's a TA?${}{}{}$ – DonAntonio Feb 1 '18 at 22:17
• @DonAntonio a teaching assistant. It means that in exchange for being able to get my PhD for free, I have to run calculus workshops for first year undergrads. – ALannister Feb 1 '18 at 22:19
• @AL Thanks, I think we call them "instructor" over here. – DonAntonio Feb 1 '18 at 22:20
• @DonAntonio regardless of what we're called and where, can you help me figure out how to do this problem? – ALannister Feb 1 '18 at 22:21

If the terms $a_{k}$ can be positive or negative, then in general $a_{n}$ might be more or less than the tail $t_{n}$, in which case the "solution" you've been given is not even correct. (edit: I've just realised they're asking about a particular sum in which the terms are positive, not sums in general, so I guess the solution is indeed correct)
If the terms $a_{k}$ are strictly positive on the other hand it's trivial:
$0 < t_{n+1}$
$a_{n} < a_{n} + t_{n+1} = t_{n}$
• we were specifically told that the terms $a_{k}$ are positive. Anyway, look at the series I provided. There's no way that any of those terms could be negative or zero. Also, for a geometric series $\sum_{k=0}^{\infty}\left(\frac{8}{9} \right)^{k}$, same idea? – ALannister Feb 1 '18 at 22:25
• Yep just saw that. So the point is for any series with positive terms $a_{n} < t_{n}$ because $t_n$ includes $a_n$ as well as a bunch of other (positive) terms – Steven Irrgang Feb 1 '18 at 22:27