Algebra (Olympiad)hardFree

Find T_n/S_n with Harmonic Sum and Catalan Identity | IOQM

JEE Maths question with a full step-by-step solution.

Question

S_n=\sum_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}\qquad\text{and}\qquad T_n=\sum_{k=1}^{n}\frac{1}{k},

then find the value of

\dfrac{T_n}{S_n}

Solution

Answer: 2

Step 1: Split the general term of

S_n

by partial fractions. Observe that the two factors in the denominator differ by exactly

k

, since

(2n-k+1)-(2n-2k+1)=k.

Writing the numerator

k

as this difference,

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

Therefore

S_n=\sum_{k=1}^{n}\frac{1}{2n-2k+1}-\sum_{k=1}^{n}\frac{1}{2n-k+1}.

Step 2: Write out each sum. As

k

runs from

1

n

, the first denominators

2n-2k+1

take the values

2n-1,\,2n-3,\,\ldots,\,3,\,1

(the odd numbers), and the second denominators

2n-k+1

take the values

2n,\,2n-1,\,\ldots,\,n+1

. Hence

S_n=\left(1+\frac13+\frac15+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right).

Step 3: Recall Catalan's identity, which expresses the second bracket as an alternating harmonic sum:

1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}.

Proof: the left side equals

\left(1+\frac12+\cdots+\frac{1}{2n}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{2n}\right)=\left(1+\frac12+\cdots+\frac{1}{2n}\right)-\left(1+\frac12+\cdots+\frac1n\right),

and the difference of these two harmonic sums is exactly

\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}

. Step 4: Substitute Catalan's identity into

S_n

, replacing the second bracket by the alternating sum:

S_n=\left(1+\frac13+\frac15+\cdots+\frac{1}{2n-1}\right)-\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2n-1}-\frac{1}{2n}\right).

The positive odd-denominator terms

1,\dfrac13,\dfrac15,\ldots,\dfrac{1}{2n-1}

appear in both brackets and cancel, leaving only the even-denominator terms (with a plus sign):

S_n=\frac12+\frac14+\frac16+\cdots+\frac{1}{2n}.

Step 5: Factor out

\dfrac12

S_n=\frac12\left(1+\frac12+\frac13+\cdots+\frac1n\right)=\frac12\,T_n.

Step 6: Therefore

\frac{T_n}{S_n}=\frac{T_n}{\tfrac12\,T_n}=2.

Answer: 2

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