# Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$

Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$

I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequality. It is a problem of Inequality.

Edit: $n$ is a positive integer such that $n>1$.

• What values does $n$ take? – fosho Feb 6 '16 at 8:14
• @Daniel positive integer greater than 1 – user1942348 Feb 6 '16 at 8:17

The sum can be written as \begin{align} \frac{1}{n+1} + \frac{1}{n+3} + \ldots + \frac{1}{3n - 1} & = \sum_{i=1}^n \frac{1}{n + 2i - 1}. \end{align} Now recall the AM-HM inequality: $$\frac 1n\sum_{i=1}^n(n + 2i - 1) > \frac{n}{\sum_{i=1}^n \frac{1}{n + 2i - 1}}.$$ (The requirement that $n > 1$ guarantees that the inequality is strict.)

Rearrange to get \begin{align} \sum_{i=1}^n \frac{1}{n + 2i - 1} & > \frac{n^2}{\sum_{i=1}^n(n + 2i - 1)} = \frac 12. \end{align}

• How do you understand that there are n terms in the series $n+1, n+3, n+5, \cdots 3n-1$? – user1942348 Feb 6 '16 at 18:52
• @user1942348 You can notice that the step size is 2, the first number is $n + 1$, and the last number is $3n - 1$. Therefore, the number of terms is $\frac{(3n - 1) - (n + 1)}{2} + 1 = n$. You can also test by substitution: What should $i$ be for $n + 2i - 1$ to be equal to $n + 1$ and $3n - 1$? The answer is $1$ and $n$. – Tunococ Feb 7 '16 at 1:48

$f(x) = 1/x$ is strictly convex, therefore $$\frac{1}{2n} < \frac 12 \left( \frac{1}{n+k} + \frac{1}{3n-k} \right)$$ for $k = 1, ..., n-1$, or $$\frac{1}{n+k} + \frac{1}{3n-k} > \frac {1}{2n} + \frac {1}{2n}$$ Combining terms pairwise from both ends of the sum shows that $$\frac{1}{n+1} + \frac{1}{n+3}+\dots+\frac{1}{3n-3} + \frac{1}{3n-1} > \underbrace{\frac {1}{2n} + \frac {1}{2n} + \dots +\frac {1}{2n} + \frac {1}{2n}}_{n \text{ terms}} = \frac 12.$$ (If $n$ is odd then the middle term $\frac {1}{2n}$ is not combined with another one. But since $n> 1$ there is at least one "pair" to combine, which gives the strict inequality.)

Any statement that needs to be proved for all $n\in\mathbb{N}$, will need to make use of induction at some point.

We have \begin{align} S_n & = \sum_{k=1}^n \dfrac1{n+2k-1} = \dfrac12 \left(\sum_{k=1}^n \dfrac1{n+2k-1} + \underbrace{\sum_{k=1}^n \dfrac1{3n-2k+1}}_{\text{Reverse the sum}}\right)\\ & = \dfrac12 \sum_{k=1}^n \dfrac{4n}{(n+2k-1)(3n-2k+1)} = \sum_{k=1}^n \dfrac{2n}{(n+2k-1)(3n-2k+1)} \end{align} From AM-GM, we have $$4n = (n+2k-1) + (3n-2k+1) \geq 2 \sqrt{(n+2k-1)(3n-2k+1)}$$ This gives us that $$\dfrac1{(n+2k-1)(3n-2k+1)} \geq \dfrac1{4n^2}$$ Hence, we obtain that $$S_n = \dfrac12 \sum_{k=1}^n \dfrac{4n}{(n+2k-1)(3n-2k+1)} \geq \sum_{k=1}^n \dfrac{2n}{4n^2} = \dfrac12$$

Also, just to note, every step in the above solution requires induction.

Also, as @MartinR rightly points out, the inequality is strictly in our case for almost all $k$ except for $k=\dfrac{n+1}2$ (since equality holds only when $n+2k-1 = 3n-2k+1 \implies k = \dfrac{n+1}2$).

• "Any statement that needs to be proved for all $n\in\mathbb{N}$, will need to make use of induction at some point". Very true and apparently very misunderstood by most people. You can't escape induction, period. – Git Gud Feb 6 '16 at 8:48
• You might add that there is a strict inequality in your AM-GM inequality for all $k$ with at most one exception, leading to the strict inequality that is asked for in the question. – Martin R Feb 6 '16 at 9:12

One approach: There are

$$\frac{(3n-1)-(n+1)}{2} + 1 = n$$ terms.

Hence if we rewrite the sum as

$$\frac 1n \sum_{j=1}^n \frac{1}{1 + (2j - 1)/n}$$

we have something that looks like a Riemann upper sum for some definite integral. Use that integral as a lower bound for the sum.

Let $$a_n = \frac{1}{n+1}+\frac{1}{n+3}+\ldots+\frac{1}{3n-1}.$$ We have $a_1=\frac{1}{2}$ and: $$\begin{eqnarray*} a_{n+2}-a_n &=& \frac{1}{3n+5}+\frac{1}{3n+3}+\frac{1}{3n+1}-\frac{1}{n+1}\\&>&\frac{3}{3n+3}-\frac{1}{n+1} = 0\end{eqnarray*}$$ so the claim is trivial, since $a_2>a_1$ and $a_{n+2}>a_n$.