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

$$\mathrm{if}{\mathrm{S}}_{\mathbf{n}}\left(\mathrm{t}\right)={\mathrm{n}}^{1-\mathbf{t}}(\frac{{(\mathrm{n}+1)}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}+1]{(\mathrm{n}+1)!}\right)}^{\mathbf{t}}}-\frac{{\mathrm{n}}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}]{\mathrm{n}!}\right)}^{\mathbf{t}}})$$$$\mathrm{with}\mathrm{t}>0$$$$\text{Double subscripts: use braces to clarify}$$

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

$${\mathrm{S}}_{\mathbf{n}}\left(\mathrm{t}\right)={\mathrm{n}}^{1-\mathbf{t}}(\frac{{(\mathrm{n}+1)}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}+1]{(\mathrm{n}+1)!}\right)}^{\mathbf{t}}}-\frac{{\mathrm{n}}^{2\mathbf{t}}}{{\left(\sqrt[\mathbf{n}]{\mathrm{n}!}\right)}^{\mathbf{t}}})$$$$S_n\left(t\right)=n^{1-t}(\frac{{(n+1)}^{2t}}{{\left(\sqrt[n+1]{\sqrt{2\pi (n+1){\left(\frac{n+1}{e}\right)}^{n+1}}}\right)}^t}-\frac{n^{2t}}{{\left(\sqrt[n]{\sqrt{2\pi n{\left(\frac{n}{e}\right)}^n}}\right)}^t})$$$$=\frac{1}{n^{t-1}}(\frac{{(n+1)}^{2t}}{{\left(2\pi e^{-1}{(n+1)}^{n+2}\right)}^{\frac{t}{2(n+1)}}}-\frac{n^{2t}}{{\left(2\pi e^{-1}{n}^{n+1}\right)}^{\frac{t}{2n}}})$$$$=\frac{1}{n^{t-1}}(\frac{{(n+1)}^{2t-\frac{t(n+2)}{2(n+1)}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2(n+1)}}}-\frac{n^{2t-\frac{t(n+1)}{2n}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2n}}})$$$$=\frac{1}{n^{t-1}}(\frac{{(n+1)}^{\frac{3tn+2t}{2(n+1)}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2(n+1)}}}-\frac{n^{\frac{3tn-t}{2n}}}{{\left(2\pi e^{-1}\right)}^{\frac{t}{2n}}})$$

Commented by mathdanisur last updated on 19/Sep/21

$$\mathrm{Thank}\mathrm{You}\mathbf{S}\mathrm{er},\mathrm{answer}=1$$

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