Proof of Inequalities Involving Series or Sequence Terms

The question asks to establish a specific inequality or bound involving terms of a sequence, partial sums, or series-related expressions, not primarily about convergence.

isi-entrance 2026 QB3 View
Define $S _ { n } = \frac { 1 } { 2 } \cdot \frac { 3 } { 4 } \cdots \cdot \frac { 2 n - 1 } { 2 n }$ where $n$ is a positive integer. Then
(A) $S _ { n } < \frac { 1 } { \sqrt { 4 n + 2 } }$ for some $n > 2$.
(B) $S _ { n } < \frac { 1 } { \sqrt { 2 n + 1 } }$ for all $n \geq 2$.
(C) $S _ { n } < \frac { 1 } { \sqrt { 2 n + 5 } }$ for all $n \geq 2$.
(D) $S _ { n } > \frac { 1 } { \sqrt { 4 n + 2 } }$ for all $n \geq 2$.
jee-advanced 2008 Q9 View
Let $$S _ { n } = \sum _ { k = 1 } ^ { n } \frac { n } { n ^ { 2 } + k n + k ^ { 2 } } \quad \text { and } \quad T _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { n } { n ^ { 2 } + k n + k ^ { 2 } } ,$$ for $n = 1,2,3 , \cdots$. Then,
(A) $\quad S _ { n } < \frac { \pi } { 3 \sqrt { 3 } }$
(B) $\quad S _ { n } > \frac { \pi } { 3 \sqrt { 3 } }$
(C) $T _ { n } < \frac { \pi } { 3 \sqrt { 3 } }$
(D) $T _ { n } > \frac { \pi } { 3 \sqrt { 3 } }$