Question

The value of $\frac{\sum\limits_{r=1}^{n} \frac{1}{r}}{\sum\limits_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}}$ is

Answer 1

2

Answer 2

Solution:

We are given

$$S = \frac{\sum_{r=1}^{n} \frac{1}{r}}{\sum_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}}.$$

Let

$$H_n=\sum_{r=1}^{n}\frac{1}{r}.$$

Focus on the denominator:

$$D = \sum_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}.$$

Step 1. Replace $k$ by $n-j,$ where $j = 0,1,2,\ldots,n-1.$ Then:

$$2n-2k+1 = 2n-2(n-j)+1 = 2j+1,$$ $$2n-k+1 = 2n-(n-j)+1 = n+j+1.$$

Thus,

$$D = \sum_{j=0}^{n-1}\frac{n-j}{(2j+1)(n+j+1)}.$$

Step 2. Write the general term in partial fractions:

$$\frac{n-j}{(2j+1)(n+j+1)} = \frac{A}{2j+1}+\frac{B}{n+j+1}.$$

Multiplying both sides by $(2j+1)(n+j+1)$ gives:

$$n-j = A(n+j+1)+B(2j+1).$$

Comparing coefficients:

• Coefficient of $j$: $A+2B=-1$.
• Constant term: $A(n+1)+B = n$.

Solving $A+2B=-1$ gives $A = -1-2B.$ Substitute into the constant term:

$$(-1-2B)(n+1)+B = n \quad \Longrightarrow \quad -(n+1)-2B(n+1)+B = n,$$ $$-(n+1)-B(2n+2-1) = n,$$ $$-(n+1)- (2n+1)B = n.$$

Thus,

$$(2n+1)B = - (n+1+n) = - (2n+1) \quad \Longrightarrow \quad B=-1.$$

Then, $A=-1-2(-1)=1.$

So,

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

Step 3. Substitute back in the sum:

$$D = \sum_{j=0}^{n-1}\left[\frac{1}{2j+1} - \frac{1}{n+j+1}\right] = \underbrace{\sum_{j=0}^{n-1}\frac{1}{2j+1}}_{S_1} - \underbrace{\sum_{j=0}^{n-1}\frac{1}{n+j+1}}_{S_2}.$$

Note that $S_2 = \sum_{m=n+1}^{2n}\frac{1}{m}=H_{2n}-H_{n}.$

Also, $S_1$ is the sum of reciprocals of the first $n$ odd numbers:

$$S_1 = \frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{2n-1} = H_{2n} - \frac{1}{2}H_n,$$

since the sum of reciprocals for even numbers (up to $2n$) is $\frac{1}{2}H_n$.

Thus,

$$D = \left(H_{2n}-\frac{H_n}{2}\right) - (H_{2n}-H_n)= H_n -\frac{H_n}{2} = \frac{H_n}{2}.$$

Step 4. Now, the overall expression becomes:

$$S = \frac{H_n}{\frac{H_n}{2}} = 2.$$