evaluate $\lim_{n\to\infty}\frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}}$

Question

How do I evaluate this limit?

$$\lim_{n\to\infty}\frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}}$$

We know that $$\sum_{i=1}^{n}\frac{1}{i}\approx\ln(n)+\gamma$$

where the value of $\gamma$ is $0.577$

Similarly we can also say that $$\sum_{i=0}^{n}\frac{1}{i+1}=\ln(n+1)+\gamma$$

Therefore we can again say that $$\sum_{i=0}^{n+3}\frac{1}{i+1}=\ln(n+4)+\gamma$$

Again we can say that $$\sum_{i=0}^{n+7}\frac{1}{2i+1}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2n+15}$$

But after this step I can't find the exact value of the limit.

Wolfram Alpha is giving the result as $2$.

Hint: the numerator is $H_{n+4}\sim\ln n$ and the denominator is $H_{2n+15}-\frac12H_{n+7}\sim\ln n-\frac12\ln n$. — Anne Bauval, Dec 18 '23 at 19:10
Sorry to point it out. But you have not added the $\gamma$ sign.@AnneBauval. As far as I know the formula is $\ln(n)+\gamma$ — Syamaprasad Chakrabarti, Dec 18 '23 at 19:14
Please review the definition of $\sim$. If $x_n\to\infty$ then $x_n+\gamma\sim x_n.$ I also used for instance that $\ln(2n+15)\sim\ln n$. — Anne Bauval, Dec 18 '23 at 19:17
I have understood the way by which you are trying to solve@AnneBauval. Since $n\to\infty$ therefore $n+4\to n$. — Syamaprasad Chakrabarti, Dec 18 '23 at 19:21
You have two growing sequences: $a_n=\sum_{i=0}^{n+3}\frac{1}{i+1}$ and $b_n=\sum_{i=0}^{n+7}\frac{1}{2i+1}$. Stolz–Cesàro theorem is applicable -https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem — Svyatoslav, Dec 18 '23 at 19:22
@Anne Bauval, we are in the conditions of the theorem; therefore,$$L=\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\frac1{n+5}}{\frac1{2(n+8)+1}}$$ — Svyatoslav, Dec 18 '23 at 19:37
Two very similar problems: https://math.stackexchange.com/q/3791827/42969, https://math.stackexchange.com/q/2816227/42969. — Martin R, Dec 19 '23 at 08:54

score 7 · Answer 1 · answered Dec 18 '23 at 20:05

We have $$ \frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}} \leq \frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+2}} = 2\left(1-\frac{\sum_{i=n+4}^{n+7}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{i+1}}\right) \rightarrow2 $$ and $$ \frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}} = \frac{\sum_{i=1}^{n+4}\frac{1}{i}}{1+\sum_{i=1}^{n+7}\frac{1}{2i+1}} \geq \frac{\sum_{i=1}^{n+4}\frac{1}{i}}{1+\sum_{i=1}^{n+7}\frac{1}{2i}} = 2\left(1-\frac{2+\sum_{i=n+5}^{n+7}\frac{1}{i}}{2+\sum_{i=1}^{n+7}\frac{1}{i}}\right) \rightarrow2 $$

as $n\rightarrow \infty$.

Hence $$ \lim_{n\to\infty}\frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}}=2 $$ follows.

score 7 · Answer 2 · answered Dec 18 '23 at 20:52

7

As noticed in the comments by Stolz-Cesaro we have

$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\sum_{i=0}^{n+4}\frac{1}{i+1}-\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+8}\frac{1}{2i+1}-\sum_{i=0}^{n+7}\frac{1}{2i+1}}= \frac{\frac{1}{n+5}}{\frac{1}{2n+17}}=\frac{2n+17}{n+5} \to 2$$

which implies $$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}}=2$$

answered Dec 18 '23 at 20:52

user

154,566

Because $\frac ab=\frac cd=\frac{a-c}{b-d}$? – Bob Dobbs Dec 18 '23 at 22:43
@BobDobbs Your claim seems true but I’m sorry, I can’t see how it is related to this question. – user Dec 18 '23 at 22:53
2

Quick proof of Stolz-Cesaro theorem.:) – Bob Dobbs Dec 18 '23 at 23:08
Ah ok! Very quick indeed ;) – user Dec 18 '23 at 23:17

Mark Viola · Answer 3 · 2023-12-18T20:36:25.823

It is not true that $\sum_{i=1}^n \frac1i = \log(n) +\gamma$. Instead, it is true that $\sum_{i=1}^n \frac1i = \log(n) +\gamma+O\left(\frac1n\right)$. This can be shown using, for example, the Euler-Maclaurin Summation Formula.

Using this precise expression, we note that for the denominator

$$\begin{align} \sum_{i=0}^{n+7}\left(\frac1{2i+1}\right)&=\sum_{i=1}^{2n+15}\frac1i-\sum_{i=1}^{n+7}\frac1{2i}\\\\ &=\left(\log(2n+15)+\gamma +O\left(\frac1{2n+15}\right)\right)\\\\ &-\frac12\left(\log(n+7)+\gamma +O\left(\frac1{n+7}\right)\right)\\\\ &=\frac12 \log(n)+\frac12 \gamma+\log(2)+O\left(\frac1n\right) \end{align}$$

while for the numerator

$$\begin{align} \sum_{i=0}^{n+3}\left(\frac1{i+1}\right)&=\log(n+4)+\gamma+O\left(\frac1{n+4}\right)\\\\ &=\log(n)+\gamma+O\left(\frac1{n}\right) \end{align}$$

Can you finish now?

Please let me know how I can improve my answer. I really want to give you the best answer I can. — Mark Viola, Dec 22 '23 at 14:28

score 3 · Answer 4 · answered Dec 18 '23 at 20:39

Let $F(n)=\sum_{i=0}^{n+3}\frac{1}{i+1}, G(n)=\sum_{i=0}^{n+7}\frac{1}{2i+1}.$

We have $\int_0^{n+3}\frac{dx}{x+1}<F(n)<1+\int_0^{n+3}\frac{dx}{x+1}\tag1$ and $\int_0^{n+7}\frac{dx}{2x+1}<G(n)<1+\int_0^{n+7}\frac{dx}{2x+1}\tag2.$ From $(1)$ and $(2)$ $\frac{\ln(n+4)}{\frac12\ln(2n+15)}<\frac{F(n)}{G(n)}<\frac{1+\ln(n+4)}{\frac12\ln(2n+15)}.\tag3$ Now apply Squeeze theorem in $(3).$

score 3 · Answer 5 · answered Dec 19 '23 at 05:45

Using harmonic numbers

$$\sum_{i=0}^{n+3}\frac{1}{i+1}=H_{n+4}-1$$ $$\sum_{i=0}^{n+7}\frac{1}{2i+1}=\frac{1}{2}H_{\frac{2n+15}{2}}+\log (2)-1$$

Using the asymptotics $$H_p=\log (p)+\gamma +\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$ and simplifying

$$\frac{\sum_{i=0}^{n+3}\frac{1}{i+1} } {\sum_{i=0}^{n+7}\frac{1}{2i+1} }=2-\frac{2 (2 \log (2)-1)}{\log (4 n)+\gamma -2}+\cdots$$

For $n=1000$, the exact value is $1.88668$ while the above (highly truncated) approximation gives $1.88756$.

evaluate $\lim_{n\to\infty}\frac{\sum_{i=0}^{n+3}\frac{1}{i+1}}{\sum_{i=0}^{n+7}\frac{1}{2i+1}}$

5 Answers5