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I have the following problem:

Evaluate $$ \lim_{n\to\infty} { {1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +\ldots+ \frac{1}{n}} \over {1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \ldots+\frac{1}{2n + 1} }} $$

I tried making it into two sums, and tried to make it somehow into an integral, but couldn't find an integral.

The sums I came up with,

$$ \lim_{n\to\infty} { \sum_{k=1}^{n} {\frac{1}{k}} \over {\sum_{k=0}^{n} {\frac{1}{2k + 1}}}} $$

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  • $\begingroup$ The top sum is a Riemann sum for $\int_1^{n+1}1/x\,dx$. The denominator is $\sum_{k=1}^{2n}\frac1k-2\sum_{k=1}^n\frac1k$, which are both also Riemann sums. $\endgroup$ – Mike Earnest Jun 11 '18 at 20:53
  • $\begingroup$ Thanks for your help! $\endgroup$ – Guysudai1 Jun 11 '18 at 20:57
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Using Stolz–Cesàro theorem we have:

$$\lim_{n\to\infty} { \sum_{k=1}^{n} {\frac{1}{k}} \over {\sum_{k=0}^{n} {\frac{1}{2k + 1}}}}=\lim_{n\to\infty} { \sum_{k=1}^{n+1} {\frac{1}{k}}-\sum_{k=1}^{n} {\frac{1}{k}} \over {\sum_{k=0}^{n+1} {\frac{1}{2k + 1}}}-{\sum_{k=0}^{n} {\frac{1}{2k + 1}}}}=\lim_{n\to\infty}\frac{2n+3}{n+1}=2$$

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  • $\begingroup$ Why does $${\sum_{k=1}^{n} \frac{1}{k} } \over {\sum_{k=0}^{n} \frac{1}{2k + 1}} $$ turn into $$ {\sum_{k=1}^{n+1} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} } \over {\sum_{k=0}^{n+1} \frac{1}{2k + 1} - \sum_{k=0} ^ {n} \frac{1}{2k+1}} $$ $\endgroup$ – Guysudai1 Jun 11 '18 at 20:54
  • $\begingroup$ @fdsaddsa The equality is for the limits. Refer to en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem $\endgroup$ – gimusi Jun 11 '18 at 20:56
  • $\begingroup$ ok Thanks! great solution. $\endgroup$ – Guysudai1 Jun 11 '18 at 20:56
  • $\begingroup$ @fdsaddsa Stolz-Cesaro is a very helpful criteria for these kind of limits with summations. $\endgroup$ – gimusi Jun 11 '18 at 20:57
  • $\begingroup$ Great to know that exists. Ill be sure to use it in problems like that $\endgroup$ – Guysudai1 Jun 11 '18 at 21:00
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Hint Denote the $n$th harmonic number by $$H_n := 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}.$$ Then, the numerator of the given ratio is $H_n$, and the denominator can be written as \begin{align*} 1 + \tfrac{1}{3} + \tfrac{1}{5} + \cdots + \tfrac{1}{2 n + 1} &= \left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{2 n}\right) - \left(\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \cdots + \tfrac{1}{2 n}\right) + \tfrac{1}{2 n + 1} \\ &= \left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{2 n}\right) - \tfrac{1}{2}\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{n}\right) + \tfrac{1}{2 n + 1} \\ &= H_{2 n} - \tfrac{1}{2} H_{n} + \frac{1}{2 n + 1} . \end{align*} Now, using appropriate Riemann sum estimates gives that $$H_n = \log n + O(1).$$

Additional hint So, the denominator is $$\log (2 n) - \tfrac{1}{2} \log n + O(1) = \tfrac{1}{2} \log n + O(1),$$ and so the ratio is $$\frac{\log n}{\tfrac{1}{2} \log n} + O((\log n)^{-1}) = 2 + O((\log n)^{-1}) .$$

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