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Hello everyone how can I calculate the limit of:

$$\lim _{n\to \infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}}\right)?$$

I tried to convert this to something that looks like Riemann sum $$\lim _{n\to \infty}\left(\frac{\sum^n_{k=0}(\frac{1}{k})}{1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}}\right).$$

But I don't know how to continue.

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    $\begingroup$ Hint: The numerator is $H_n$ and the denominator is $H_{2n+1}-\frac12H_n,$ where $H_n$ is the harmonic series. We know a lot about how to approximate these values. $\endgroup$ – Thomas Andrews Jun 26 at 13:19
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    $\begingroup$ It's a shame it's not $$\lim _{n\to \infty }\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}}\right)$$ $\endgroup$ – Angina Seng Jun 26 at 13:19
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By Stolz-Cesaro we have: $\lim\limits_{n \to \infty} \frac{1 + \frac12 + \dots + \frac1n}{1 + \frac13 + \frac15 + \dots + \frac1{2n+1}} = \lim\limits_{n \to \infty} \frac{\frac1{n+1}}{\frac1{2n + 3}} = \lim\limits_{n \to \infty} \frac{2n+3}{n+1} = 2$

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  • $\begingroup$ Thank you very much!!! $\endgroup$ – xxx Jun 26 at 14:14
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One approach is as follows: it suffices to note that $$ \sum_{k=2}^n\frac{1}{k} \leq \int_1^n \frac 1x \,dx \leq \sum_{k=1}^n\frac{1}{k}, \\ \sum_{k=2}^{n+1}\frac{1}{2k-1} \leq \int_1^{n+1} \frac 1{2x-1} \,dx \leq \sum_{k=1}^{n+1}\frac{1}{2k-1}, $$ and apply the squeeze theorem. In particular, we can use the above to get $$ \frac{\ln(n)}{1 + \frac 12 \ln(2n + 1)} \leq \frac{1+\frac{1}{2}+\frac{1}{3}+…\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1}} \leq \frac{1 + \ln(n)}{\frac 12 \ln(2n + 1)}. $$


Another approach: note that adding a final $\frac 1{2n + 2}$ to $1/2$ times the numerator yields $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}$, and $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2} \leq \\ 1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1} \leq \\ 1 + \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}\right). $$

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    $\begingroup$ The sums on the left and right side are equal in both lines. That can’t be what you mean. $\endgroup$ – Thomas Andrews Jun 26 at 13:25
  • $\begingroup$ @ThomasAndrews thanks for catching that $\endgroup$ – Omnomnomnom Jun 26 at 13:27
  • $\begingroup$ The second integral is $\log(2n-1).$ $\endgroup$ – Thomas Andrews Jun 26 at 13:32
  • $\begingroup$ @ThomasAndrews Yet another mistake we caught around the same time. Should be all set now. $\endgroup$ – Omnomnomnom Jun 26 at 13:34
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\begin{align*} \text{Required limit}&=\lim_{n\to\infty}\left(\dfrac{\sum\limits_{i=1}^n\dfrac1i}{\sum\limits_{i=1}^{2n+1}\dfrac1i-\sum\limits_{i=1}^n\dfrac1{2i}}\right)\\ &=\lim_{n\to\infty}\left(\dfrac{H_n}{H_{2n+1}-\dfrac12H_n}\right)\\ &=\lim_{n\to\infty}\left(\dfrac{\dfrac{H_n}{\log(2n+1)}}{\quad\dfrac{H_{2n+1}}{\log(2n+1)}-\dfrac{H_n}{2\log(2n+1)}\quad}\right)\\ &=\dfrac{1}{1-\dfrac12}\\ &=\boxed2 \end{align*}

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  • $\begingroup$ I didn't understand the jump between the third and last steps. $\endgroup$ – xxx Jun 26 at 14:00
  • $\begingroup$ @xxx, I am using the fact that $H_n\sim\log n$. $\endgroup$ – Martund Jun 26 at 14:50
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$$H_n = \log n + O(1)$$ so $$H_{2n+1} - \tfrac{1}{2} H_n = \log(2n + 1) - \tfrac{1}{2} \log n + O(1) = \tfrac{1}{2} \log n + O(1)$$ whence $$\frac{H_n}{H_{2n+1} - \tfrac{1}{2}H_n} = \frac{\log n + O(1)}{\tfrac{1}{2}\log n + O(1)} \to 2.$$

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$ {1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}}+\cdots=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots) = \frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots) $ $$ \lim _{n\to \infty \:}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}}\right)=\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots\frac{1}{n}+\cdots}{\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\cdots)}\right)=2 $$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Jun 27 at 13:33
  • $\begingroup$ @AloizioMacedo : Actually the mathjax is now easier to copy and paste, thank you for moving it to chat $\endgroup$ – jimjim Jun 28 at 1:03

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