$S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$.
Prove that $\dfrac{29}{27}<S<\dfrac{7}{6}$.
My Attempt:
$S<\dfrac{500}{1000} + \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$
(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)
Using AM-HM inequality gave me $S > 1$, but the bounds are weak.
Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$
Inequality with sum of inverses of consecutive numbers
The answers to these questions are nice, but the bounds are weak.
Any help without calculus and without calculations involving calculators would be appreciated.
(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))
Best Answer
Upper bound $\dfrac{7}{6}$
Your question describes a method of breaking the range up into sub-ranges, and then estimating the sum over a sub-range by noting that each term is $\le$ the first term. But we can do much better if we estimate this sum using a linear approximation instead. Then we only need three sub-ranges to establish the upper bound of $\frac76$.
Consider a 'graph' of the function $$f(k)=\frac{1}{k}$$ for integers $k\ge 1$. If we look at any interval $m\le k\le n$ with $1\le m\le n-2$, we can compare it with the straight-line function joining points $(m,\frac{1}{m})$ and $(n,\frac{1}{n})$. We can define this function explicitly if you want, as: $$g(k)=\frac{1}{n-m}\left(\frac{k-m}{n}+\frac{n-k}{m}\right)$$ Then $f(m)=g(m),f(n)=g(n)$; and for $m<k<n,f(k)<g(k)$ (because $f$ is a convex function).
So $$\sum_{k=m}^nf(k)<\sum_{k=m}^ng(k)$$ But the $g(k)$ are in arithmetic progression, so the RHS is half the sum of the first and the last terms multiplied by the number of terms, i.e. $$\frac{n-m+1}{2}\left(\frac{1}{m}+\frac{1}{n}\right)$$
We divide the range into three equal sub-ranges of size $667$. This gives us: $$\begin{align} \sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\ & <\frac{667}{2}\left(\frac{1}{1001}+\frac{1}{1667}\right)+\frac{667}{2}\left(\frac{1}{1668}+\frac{1}{2334}\right)+\frac{667}{2}\left(\frac{1}{2335}+\frac{1}{3001}\right)\\ & <\frac{667}{2}\left(\frac{1}{1000.5}+\frac{1}{1667.5}+\frac{1}{1667.5}+\frac{1}{2334.5}+\frac{1}{2334.5}+\frac{1}{3001.5}\right)\\ & =\frac13+\frac15+\frac15+\frac17+\frac17+\frac19\\ & =\frac{356}{315}\\ & <\frac{7}{6}\end{align}$$
where we have used the fact that $\frac{1}{m}+\frac{1}{n}<\frac{1}{m-\frac12}+\frac{1}{n+\frac12}$ if $m<n$, again by convexity of $f(k)$.
Note that if the last term were $\frac{1}{3000}$, we could do this with just two sub-ranges, and we would get exactly $\frac{7}{6}$ as a bound. So perhaps I have missed an easier way.
Lower bound $\dfrac{29}{27}$
This time we use the convexity of the function to estimate the sum over a sub-range from below. The sum of $\frac{1}{k}$ over the range $m\le k\le n$ is greater than the number of elements multiplied by the middle element if the range contains an odd number of elements. So we get $$\begin{align} \sum_{k=1001}^{3001}\frac{1}{k} & =\sum_{k=1001}^{1667}\frac{1}{k}+\sum_{k=1668}^{2334}\frac{1}{k}+\sum_{k=2335}^{3001}\frac{1}{k}\\ & > 667\left(\frac{1}{1334}+\frac{1}{2001}+\frac{1}{2668}\right)\\ & = \frac12+\frac13+\frac14\\ & = \frac{13}{12}\\ & >\frac{29}{27} \end{align}$$