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Question Number 154495 by mathdanisur last updated on 18/Sep/21

if S_{n} (t) = n^{1 - t} (\frac{{(n + 1)}^{2 t}}{{(\sqrt[n + 1]{(n + 1)!})}^{t}} - \frac{n^{2 t}}{{(\sqrt[n]{n!})}^{t}})

with t > 0

Double subscripts: use braces to clarify

Answered by Ar Brandon last updated on 18/Sep/21

S_{n} (t) = n^{1 - t} (\frac{{(n + 1)}^{2 t}}{{(\sqrt[n + 1]{(n + 1)!})}^{t}} - \frac{n^{2 t}}{{(\sqrt[n]{n!})}^{t}})

S_{n} (t) = n^{1 - t} (\frac{{(n + 1)}^{2 t}}{{(\sqrt[n + 1]{\sqrt{2 π (n + 1) {(\frac{n + 1}{e})}^{n + 1}}})}^{t}} - \frac{n^{2 t}}{{(\sqrt[n]{\sqrt{2 π n {(\frac{n}{e})}^{n}}})}^{t}})

= \frac{1}{n^{t - 1}} (\frac{{(n + 1)}^{2 t}}{{(2 π e^{- 1} {(n + 1)}^{n + 2})}^{\frac{t}{2 (n + 1)}}} - \frac{n^{2 t}}{{(2 π e^{- 1} n^{n + 1})}^{\frac{t}{2 n}}})

= \frac{1}{n^{t - 1}} (\frac{{(n + 1)}^{2 t - \frac{t (n + 2)}{2 (n + 1)}}}{{(2 π e^{- 1})}^{\frac{t}{2 (n + 1)}}} - \frac{n^{2 t - \frac{t (n + 1)}{2 n}}}{{(2 π e^{- 1})}^{\frac{t}{2 n}}})

= \frac{1}{n^{t - 1}} (\frac{{(n + 1)}^{\frac{3 t n + 2 t}{2 (n + 1)}}}{{(2 π e^{- 1})}^{\frac{t}{2 (n + 1)}}} - \frac{n^{\frac{3 t n - t}{2 n}}}{{(2 π e^{- 1})}^{\frac{t}{2 n}}})

Commented by mathdanisur last updated on 19/Sep/21

Thank You S er, answer = 1