n-1-1-n-n-1-n-please-help-me-

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

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

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

l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{k k^{\frac{1}{k}}} t h e s e q u e n c e u_{k} = \frac{1}{k . k^{\frac{1}{k}}} i s d e c r e a s i n g \Rightarrow

\int_{k}^{k + 1} \frac{d t}{t t^{\frac{1}{t}}} ⩽ f (k) ⩽ \int_{k - 1}^{k} \frac{d t}{t t^{\frac{1}{t}}} \Rightarrow \sum_{k = 2}^{n} \int_{k}^{k + 1} \frac{d t}{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{d t}{{t t}^{\frac{1}{t}}} \Rightarrow

\int_{2}^{n + 1} \frac{d t}{t . t^{\frac{1}{t}}} ⩽ S_{n} - 1 ⩽ \int_{1}^{n} \frac{d t}{t t^{\frac{1}{t}}} \Rightarrow 1 + \int_{2}^{+ \infty} \frac{d t}{t . t^{\frac{1}{t}}} ⩽ {l i m}_{n \to + \infty} S_{n} ⩽ 1 + \int_{1}^{+ \infty} \frac{d t}{t . t^{\frac{1}{t}}}

b u t \int_{1}^{+ \infty} \frac{d t}{t t^{\frac{1}{t}}} = \int_{1}^{+ \infty} \frac{d t}{t^{1 + \frac{1}{t}}} = \int_{1}^{+ \infty} \frac{d t}{e^{(1 + \frac{1}{t}) l n (t)}}

= \int_{1}^{+ \infty} e^{- (1 + \frac{1}{t}) l n (t)} d t c h a n g e m e n t t = \frac{1}{x} g i v e

\int_{1}^{+ \infty} e^{- (1 + \frac{1}{t}) l n (t)} d t = - \int_{0}^{1} e^{(1 + x) l n (x)} (- \frac{d x}{x^{2}}) = \int_{0}^{1} \frac{e^{(1 + x) l n (x)}}{x^{2}} d x

= \int_{0}^{1} \frac{x x^{x}}{x^{2}} d x = \int_{0}^{1} x^{x - 1} d x \dots . b e c o n t i n u e d \dots