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I have the following problem:

Evaluate $$ \lim_{n\to\infty}{{1+\frac12+\frac13 +\frac14+\ldots+\frac1n}\over{1+\frac13 +\frac15+\frac17+\ldots+\frac1{2n+1}}} $$

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,

$$ \lim_{n\to\infty} { \sum_{k=1}^n {\frac1k} \over {\sum_{k=0}^n {\frac{1}{2k+1}}}} $$

PinkyWay
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Guysudai1
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  • 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. – Mike Earnest Jun 11 '18 at 20:53
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    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. – Thomas Andrews Jun 26 '20 at 13:19
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    It's a shame it's not $$\lim _{n\to \infty }\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}}\right)$$ – Angina Seng Jun 26 '20 at 13:19
  • If somebody wonders why there are "duplicate answers", it was a merge. – quid May 22 '21 at 12:41

6 Answers6

17

Using Stolz–Cesàro theorem we have:

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

amWhy
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user
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  • Why does $${\sum_{k=1}^{n} \frac{1}{k} } \over {\sum_{k=0}^{n} \frac{1}{2k + 1}} $$ turn into $$ {\sum_{k=1}^{n+1} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} } \over {\sum_{k=0}^{n+1} \frac{1}{2k + 1} - \sum_{k=0} ^ {n} \frac{1}{2k+1}} $$ – Guysudai1 Jun 11 '18 at 20:54
  • @fdsaddsa The equality is for the limits. Refer to https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – user Jun 11 '18 at 20:56
  • This is good. My instincts would have made me use the asymptotic estimates of the harmonic sums :-) – Jyrki Lahtonen May 20 '21 at 03:57
  • @Guysudai1 I will add that Stolz-Cesaro is sometimes formulated for limits, sometimes for sums - both formulations are equivalent. See also this answer. – Martin Sleziak May 20 '21 at 06:34
17

By Stolz-Cesaro we have: $\lim\limits_{n \to \infty} \frac{1 + \frac12 + \dots + \frac1n}{1 + \frac13 + \frac15 + \dots + \frac1{2n+1}} = \lim\limits_{n \to \infty} \frac{\frac1{n+1}}{\frac1{2n + 3}} = \lim\limits_{n \to \infty} \frac{2n+3}{n+1} = 2$

PinkyWay
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Joitandr
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7

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 $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2} \leq \\ 1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1} \leq \\ 1 + \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}\right). $$

Ben Grossmann
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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 $$\frac{\log n}{\tfrac{1}{2} \log n} + O((\log n)^{-1}) = 2 + O((\log n)^{-1}) .$$

Travis Willse
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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*}

Martund
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2

$$H_n = \log n + O(1)$$ 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 $$\frac{H_n}{H_{2n+1} - \tfrac{1}{2}H_n} = \frac{\log n + O(1)}{\tfrac{1}{2}\log n + O(1)} \to 2.$$

Unit
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