lim-x-1-1-2-1-3-1-n-1-1-3-1-5-1-2n-1-

Question Number 101239 by M±th+et+s last updated on 01/Jul/20

\lim_{x \to \infty} (\frac{1 + \frac{1}{2} + \frac{1}{3} + \dots \dots + \frac{1}{n}}{1 + \frac{1}{3} + \frac{1}{5} \dots \dots + \frac{1}{2 n + 1}})

Answered by mathmax by abdo last updated on 01/Jul/20

1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{n} = H_{n}

1 + \frac{1}{3} + \frac{1}{5} + \dots . + \frac{1}{2 n + 1} = 1 + \frac{1}{2} + \frac{1}{3} + \dots . . + \frac{1}{2 n} + \frac{1}{2 n + 1} - \frac{1}{2} - \frac{1}{4} - \dots - \frac{1}{2 n}

= H_{2 n + 1} - \frac{1}{2} H_{n} \Rightarrow q_{n} = \frac{H_{n}}{H_{2 n + 1} - \frac{H_{n}}{2}} = \frac{1}{\frac{H_{2 n + 1}}{H_{n}} - \frac{1}{2}}

we have \frac{H_{2 n + 1}}{H_{n}} = \frac{\ln (2 n + 1) + γ + o (\frac{1}{2 n + 1})}{\ln (n) + γ + o (\frac{1}{n})}

= \frac{\ln (2 n + 1)}{\ln (n)} \times \frac{1 + \frac{γ}{\ln (2 n + 1)} + o (\frac{1}{(2 n + 1) \ln (2 n + 1)})}{1 + \frac{γ}{\ln (n)} + o (\frac{1}{nln (n)})} \Rightarrow

\lim_{n \to + \infty} \frac{H_{2 n + 1}}{H_{n}} = \lim_{n \to + \infty} \frac{\ln (2 n + 1)}{\ln (n)} = \lim_{n \to + \infty} \frac{\ln (n) + \ln (2 + \frac{1}{n})}{\ln (n)}

= \lim_{n \to + \infty} \frac{1 + \frac{\ln (2 + \frac{1}{n})}{\ln (n)}}{1} = 1 \Rightarrow \lim_{n \to + \infty} q_{n} = \frac{1}{1 - \frac{1}{2}} = 2

Commented by M±th+et+s last updated on 01/Jul/20

w e l l d o n e

Commented by mathmax by abdo last updated on 01/Jul/20

you are welcome