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Question Number 46187 by annika0209 last updated on 22/Oct/18

$$\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}\times {\mathrm{n}}^{\frac{1}{\mathrm{n}}}}=?\mathrm{please}\mathrm{help}\mathrm{me}!!!!$$$$$$

Commented by maxmathsup by imad last updated on 22/Oct/18

$$let{S}_n=\sum _{k=1}^n\frac{1}{k{k}^{\frac{1}{k}}}thesequence{u}_k=\frac{1}{k.k^{\frac{1}{k}}}isdecreasing\Rightarrow$$$${\int }_k^{k+1}\frac{dt}{t{t}^{\frac{1}{t}}}⩽f\left(k\right)⩽{\int }_{k-1}^k\frac{dt}{t{t}^{\frac{1}{t}}}\Rightarrow \sum _{k=2}^n{\int }_k^{k+1}\frac{dt}{t.t^{\frac{1}{t}}}⩽\sum _{k=2}^n\frac{1}{k.k^{\frac{1}{k}}}⩽\sum _{k=2}^n{\int }_{k-1}^k\frac{dt}{{tt}^{\frac{1}{t}}}\Rightarrow$$$${\int }_2^{n+1}\frac{dt}{t.t^{\frac{1}{t}}}⩽S_n-1⩽{\int }_1^n\frac{dt}{t{t}^{\frac{1}{t}}}\Rightarrow 1+{\int }_2^{+\mathrm{\infty }}\frac{dt}{t.t^{\frac{1}{t}}}⩽{lim}_{n\to +\mathrm{\infty }}{S}_n⩽1+{\int }_1^{+\mathrm{\infty }}\frac{dt}{t.t^{\frac{1}{t}}}$$$$but{\int }_1^{+\mathrm{\infty }}\frac{dt}{t{t}^{\frac{1}{t}}}={\int }_1^{+\mathrm{\infty }}\frac{dt}{t^{1+\frac{1}{t}}}={\int }_1^{+\mathrm{\infty }}\frac{dt}{e^{(1+\frac{1}{t})ln\left(t\right)}}$$$$={\int }_1^{+\mathrm{\infty }}{e}^{-(1+\frac{1}{t})ln\left(t\right)}dtchangementt=\frac{1}{x}give$$$${\int }_1^{+\mathrm{\infty }}{e}^{-(1+\frac{1}{t})ln\left(t\right)}dt=-{\int }_0^1{e}^{(1+x)ln\left(x\right)}(-\frac{dx}{x^2})={\int }_0^1\frac{e^{(1+x)ln\left(x\right)}}{x^2}dx$$$$={\int }_0^1\frac{x{x}^x}{x^2}dx={\int }_0^1{x}^{x-1}dx....becontinued...$$

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