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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

t_r=\frac{1^2+2^2+\cdots+r^2}{1^3+2^3+\cdots+r^3}

and

S_n=\sum_{r=1}^{n}(-1)^r t_r

, then

\lim_{n\to\infty}S_n

equals

-\frac29

-\frac13

\frac23

-\frac23

correct

Solution

Step 1: Use the standard sums for squares and cubes:

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_\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

S_\infty=\frac23(-1)=-\frac23,

Correct answer: (4)

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