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Limit of the Alternating Series of t_r Ratios | JEE

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

Question
If tr=12+22+⋯+r213+23+⋯+r3t_r=\frac{1^2+2^2+\cdots+r^2}{1^3+2^3+\cdots+r^3} and Sn=∑r=1n(−1)rtrS_n=\sum_{r=1}^{n}(-1)^r t_r, then lim⁡n→∞Sn\lim_{n\to\infty}S_n equals
A−29-\frac29
B−13-\frac13
C23\frac23
D−23-\frac23correct
Solution
Step 1: Use the standard sums for squares and cubes:
tr=r(r+1)(2r+1)6r2(r+1)24=2(2r+1)3 r(r+1)=23(1r+1r+1).t_r=\frac{\frac{r(r+1)(2r+1)}{6}}{\frac{r^2(r+1)^2}{4}}=\frac{2(2r+1)}{3\,r(r+1)}=\frac23\left(\frac1r+\frac{1}{r+1}\right).
 Step 2: So
S∞=23[(−1+12−13+⋯)+(−12+13−14+⋯)].S_\infty=\frac23\Big[\big(-1+\tfrac12-\tfrac13+\cdots\big)+\big(-\tfrac12+\tfrac13-\tfrac14+\cdots\big)\Big].
 Step 3: Add the two brackets term by term — everything cancels except the leading −1-1:
S∞=23(−1)=−23,S_\infty=\frac23(-1)=-\frac23,
 Correct answer: (4)
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