The function satisﬁes when . Explain by means of a sketch, or otherwise, why
By considering or otherwise show that, if , then
and deduce that
Deduce that as .
Noting that , show also that if then
Quite a lot of different ideas are required for this question; hence .
The ﬁrst hurdle is to decide what the constant in the ﬁrst part has to be in order to make the second part work. You might try to maximise the integrand using calculus, but that would be the wrong thing to do (you should ﬁnd if you do this that the integrand has no turning point in the given range: it increases throughout the range).
The next hurdle is the third displayed equation, which follows from the preceding result. Then there is still more work to do.
It is not at all obvious that the series (which is called the harmonic series) tends to inﬁnity as increases. There are easier ways of proving this than the method used in this question but this way tells us two interesting things. First it tells us that the sum increases very slowly indeed: the ﬁrst terms only get to 100. Second it tells us that
for all . (This is the third displayed equation in the question slightly rewritten.) Not only does the sum diverge, it does so logarithmically. In fact, in the limit ,
where is Euler’s constant. Its value is about .
23 The theorem says that the equation , where , , and are positive integers, can only hold if or
Solution to problem 31
Any sketch showing a squiggly curve all of which lies beneath the line and above the -axis will do: area under curve (the integral) is less than the area of the rectangle ().
Let’s start the next part in the obvious way, by evaluating the integral.
Noting that , we have:
The largest value of the integrand in the interval occurs at , since the numerator increases as increases and the denominator decreases as increases (remember that ). Using the result of the ﬁrst paragraph gives
Summing both sides from 2 to and cancelling lots of terms in pairs gives
Note that . Since as , so also must (it differs from the logarithm by less than 1).
Finally, rearranging the ﬁrst inequality in () gives , i.e. . Setting , and using the inequalities and (or ) gives:
This question seems very daunting at ﬁrst, because you are asked to prove a sequence of completely unfamiliar results. However, you should learn from this question that if you keep cool and follow the hints, explicit and implicit, you can achieve some surprisingly sophisticated results. You might think that this situation is very artiﬁcial: in real life, you do not receive hints to guide you. But often in mathematical research, hints are buried deep in the problem if only you can recognise them.