Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\ldots+\frac{1}{3001}<\frac43$

Prove that $$1<\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+\ldots+\dfrac{1}{3001}<\dfrac43 \, .$$

My work:
$$\begin{eqnarray*} S&=&\bigg(\dfrac{1}{1001}+\dfrac{1}{3001}\bigg)+\bigg(\dfrac{1}{1002}+\dfrac{1}{3000}\bigg)+\ldots+\dfrac{1}{2001}\\ S&=&\dfrac{1}{4002}\bigg\{\bigg(\dfrac{1001+3001}{1001}+\dfrac{1001+3001}{3001}\bigg)+\ldots\bigg\}+\dfrac{1}{2001}\\ S&\ge& \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1 \end{eqnarray*}$$

I could derive the left hand inequality with a $\ge$ sign though but could not do anything about the right hand inequality. Please help.

EDIT: I do not need to use the equality sign, I can rather use only strict inequality because there exist terms which are not equal so, the AM-HM inequality is actually a strict inequality here.

$$S> \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1$$

Though I got the solution but I am looking for some non-calculus solutions too.

• Please do not rollback to an earlier version without good reason. Note that titles are intentionally done without display math (unless it is necessary) to keep the front page as clean as possible. – TMM Feb 24 '14 at 13:33
• @TMM Sorry, but I did not see any major improvement so I rolled back. – Hawk Feb 24 '14 at 13:34
• @TMM Would you please explain what is the need to edit the post? I am asking this as I would be able to further avoid such mistakes. – Hawk Feb 24 '14 at 14:01
• I already mentioned one reason: the title was too big. Other reasons include making the maths a bit more readable, and adding another tag besides only inequality. – TMM Feb 24 '14 at 16:41

$$\log{k+1\over k}=\int_k^{k+1}{dx\over x}<{1\over k}<\int_{k-1}^k{dx\over x}=\log{k\over k-1}\qquad(k>1)\ .$$ Summing over $k$ produces telescoping series on the left hand and the right hand side. Doing the computations one finds $$\log{3002\over1001}<S<\log{3001\over 1000}\ .$$ Here the left hand side is obviously $>1$. For the right hand side we note that $$e^{4\over3}>1+{4\over3}+{(4/3)^2\over2}={29\over9}>{3001\over1000}\ ,$$ and this shows that $$\log{3001\over 1000}<{4\over3}\ .$$ Update: The OP has desired a calculus-free approach. For the upper estimate one could argue as follows: Using the splitting $$S=\sum_{k=1001}^{1350}{1\over k}+\sum_{k=1351}^{1800}{1\over k}+\sum_{k=1801}^{2400}{1\over k}+\sum_{k=2401}^{3000}{1\over k}+{1\over3001}$$ one obtains $$S<{350\over1000}+{450\over1350}+{600\over1800}+{600\over2400}+{1\over3001}={76\over60}+{1\over3001}<{4\over3}\ .$$

• how can we estimate that, $log(3)<4/3$ ? – OBDA Feb 24 '14 at 14:11
• (+1) for a very nice approach...but sir, would you please add another answer which might be non-calculus...it would be a great more help...thank you! – Hawk Feb 24 '14 at 14:39
• @Christian Blatter I think the estimation is wrong. The nominator of (4/3)^2 is 4 not 2. – OBDA Feb 24 '14 at 15:56
• It is very interesting to notice that one is able to "split and condensate", or just apply the Jensen inequality, and get an upper bound close to $\frac{7}{6}$. – Jack D'Aurizio Feb 24 '14 at 15:56
• @O.B.D.A.: For $x>0$ one has $e^x>1+x+x^2/2$. – Christian Blatter Feb 24 '14 at 16:52

Let $k$ be a fixed integer greater than $0$.

It holds that $\frac{1}{1000+k}\leq \frac{1}{1000+x}$for any $x \in [k-1,k]$.

Integrating for fixed $k \in [1,2001]$ yields $$\frac{1}{1000+k}\leq \int_{k-1}^{k}\frac{1}{1000+x}dx$$ Hence $$\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\ldots+\frac{1}{3001} \leq \int_{0}^{2001}\frac{1}{1000+x}dx=\ln\left(\frac{3001}{1000}\right)$$

And $$\ln\left(\frac{3001}{1000}\right)\sim 1.1 < \frac{4}{3}$$

For a solution without using calculus:

For the upper bound, note for integers $1 \le k \le 1000$:

$$\frac1{1000+k} + \frac1{3002-k} = \frac{4002}{(1000+k)(3002-k)}= \frac{4002}{4004001-(1001-k)^2} \le \frac{4002}{3004001}$$ where the equality is only when $k=1$.

$$\implies S = \sum_{k=1}^{1000} \left(\frac1{1000+k} + \frac1{3002-k} \right)+\frac1{2001} < \frac{4002}{3004001}\cdot 1000+\frac1{2001} = \frac{8011006001}{6011006001} < \frac43$$

For the lower bound here is another way, using $AM > HM$ for distinct numbers, we get $$\frac1{1000+k} + \frac1{3002-k}> \frac2{2001} \qquad \text{for }k = 1, 2, 3, \dots 1000$$

$$\implies S = \sum_{k=1}^{1000} \left(\frac1{1000+k} + \frac1{3002-k} \right)+\frac1{2001} > \frac{2}{2001} \cdot 1000+\frac1{2001}=1$$

I'd rather say that by the AM-HM inequality we have: $$\sum_{k=1001}^{3001}\frac{1}{k}>\frac{2001^2}{\sum_{k=1001}^{3001}k}=1\tag{1}$$ and since $\frac{1}{k}<-\log\left(1-\frac{1}{k}\right)=\log\frac{k}{k-1}$ we have: $$\sum_{k=1001}^{3001}\frac{1}{k}<\log\prod_{k=1001}^{3001}\frac{k}{k-1}=\log\frac{3001}{1000}<\log 3.\tag{2}$$ Now proving $\log 3<\frac{4}{3}$ is equivalent to prove that $e^4>27$. Since $e>\frac{5}{2}$ (by considering the first three terms of the Taylor series of the exponential function around zero, for istance), $e^4>39$, and we are fine.

Notice that by partial summation we have: $$\sum_{k=1001}^{3001}\frac{1}{k} = \frac{2001}{3001}+\sum_{k=1001}^{3000}\frac{k-1000}{k(k+1)}=\frac{2001}{3001}+\sum_{k=1}^{2000}\frac{k}{(k+1000)(k+1001)},$$ $$\sum_{k=1001}^{3001}\frac{1}{k} < \frac{2001}{3001}+\sum_{k=1}^{2000}\frac{k}{(k+1000)^2},\tag{3}$$ but since $f(x)=\frac{x}{(x+1000)^2}$ is a concave function on $I=[0,2000]$, by the Jensen inequality (or simply because $f(x)$ over $I$ is bounded by its value in $x=1000$, that is the only stationary point in $I$) we have the improved upper bound:

$$\sum_{k=1001}^{3001}\frac{1}{k}<\frac{2}{3}+\frac{1}{2}=\frac{7}{6}.\tag{4}$$

Bounding $f(x)$ between the envelope of three tangents (in $x=0,x=1000,x=2000$) and the envelope of two secants (from $x=0$ to $x=1000$ and from $x=1000$ to $x=2000$) leads to the following sharpened inequality:

$$\frac{37}{36}<\sum_{k=1001}^{3001}\frac{1}{k}<\frac{9}{8}.\tag{5}$$