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
If
Sn=∑k=1nk(2n−2k+1)(2n−k+1)andTn=∑k=1n1k,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 TnSn\dfrac{T_n}{S_n}.
Solution
Answer: 2
Step 1: Split the general term of SnS_n by partial fractions. Observe that the two factors in the denominator differ by exactly kk, since
(2n−k+1)−(2n−2k+1)=k.(2n-k+1)-(2n-2k+1)=k.
 Writing the numerator kk as this difference,
k(2n−2k+1)(2n−k+1)=12n−2k+1−12n−k+1.\frac{k}{(2n-2k+1)(2n-k+1)}=\frac{1}{2n-2k+1}-\frac{1}{2n-k+1}.
 Therefore
Sn=∑k=1n12n−2k+1−∑k=1n12n−k+1.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 kk runs from 11 to nn, the first denominators 2n−2k+12n-2k+1 take the values 2n−1, 2n−3, …, 3, 12n-1,\,2n-3,\,\ldots,\,3,\,1 (the odd numbers), and the second denominators 2n−k+12n-k+1 take the values 2n, 2n−1, …, n+12n,\,2n-1,\,\ldots,\,n+1. Hence
Sn=(1+13+15+⋯+12n−1)−(1n+1+1n+2+⋯+12n).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:
1−12+13−14+⋯+12n−1−12n=1n+1+1n+2+⋯+12n.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
(1+12+⋯+12n)−2(12+14+⋯+12n)=(1+12+⋯+12n)−(1+12+⋯+1n),\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 1n+1+1n+2+⋯+12n\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}. Step 4: Substitute Catalan's identity into SnS_n, replacing the second bracket by the alternating sum:
Sn=(1+13+15+⋯+12n−1)−(1−12+13−14+⋯+12n−1−12n).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 1,13,15,…,12n−11,\dfrac13,\dfrac15,\ldots,\dfrac{1}{2n-1} appear in both brackets and cancel, leaving only the even-denominator terms (with a plus sign):
Sn=12+14+16+⋯+12n.S_n=\frac12+\frac14+\frac16+\cdots+\frac{1}{2n}.
 Step 5: Factor out 12\dfrac12:
Sn=12(1+12+13+⋯+1n)=12 Tn.S_n=\frac12\left(1+\frac12+\frac13+\cdots+\frac1n\right)=\frac12\,T_n.
 Step 6: Therefore
TnSn=Tn12 Tn=2.\frac{T_n}{S_n}=\frac{T_n}{\tfrac12\,T_n}=2.
 Answer: 2
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