1
$\begingroup$

$a_n = \frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}$ converges to a number between $\frac{1}{2}$ and $1$. I know that it converges to Ln(2) but we still haven't learned about integrals in my class, so calculating its limits kills the purpose of the exercise.

I've been trying for a while, I tried with this theorem that states that if $a_n \leq b_n$ for all n greater or equal to some N, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$

So my aproach was to find two sequences such that their limits are $\frac{1}{2}$ and $1$, I used these $\frac{n}{2n+1}$ and $\frac{n}{n+1}$, the limit of the first one is $\frac{1}{2}$ and the limit of the second one is $1$, and we have that $$\frac{n}{2n+1} \leq \frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n} \leq \frac{n}{n+1}$$ for all $n \geq 6$, I've been trying to prove that with induction but I have been unable to do it. Is there any other way to prove that the limit is between $\frac{1}{2}$ and $1$?, or if someone could give me a hint with the induction proof, I would be very grateful.

$\endgroup$
4
$\begingroup$

Hint 1: $$a_{n+1}-a_n=\left( \frac{1}{n+1}+ \frac{1}{n+1} +\dots + \frac{1}{2n+2} \right)-\left(\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}\right)\\ = \frac{1}{2n+1}+\frac{1}{2n+2}- \frac{1}{n}<\frac{1}{2n}+\frac{1}{2n}- \frac{1}{n}=0$$

Hint 2:

$$a_n =\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n} \geq \frac{1}{2n}+ \frac{1}{2n} +\dots + \frac{1}{2n}$$

$\endgroup$
1
$\begingroup$

Let the base case be $n = 3$. Then $$1/3+1/4+1/5+1/6 = 19/20$$ is between $1/2$ and $1$. Now assuming it is true for all $n$, $$a_{n+1} = \sum_{k=n+1}^{2(n+1)}\frac{1}{k}$$ is equal to $$a_n-1/n+1/(2n+1)+1/(2n+2) = a_n - \frac{3n+2}{(2n+1)(2n+2)n} < a_n$$. This proves that $a_n$ will be less than $1$ for all $n \ge 3$.

You don't need any inductive proof to prove that $a_n$ is greater than $1/2$, Since $$\frac{1}{2} \le \sum_{k=n}^{2n}\frac{1}{2n} = \frac{n+1}{2n} \le \sum_{k=n}^{2n}\frac{1}{k}$$

Therefore, $a_n$ is between $1/2$ and $1$ for all $n \ge 3$.

Edit: Since $a_n$ is continuously decreasing and is always above $1/2$, it must converge.

$\endgroup$
  • 1
    $\begingroup$ To complete the proof, you should prove the convergence of $\{a_n\}$. $\endgroup$ – Feng Shao Jul 5 '19 at 4:57
  • $\begingroup$ I edited the answer. $\endgroup$ – Varun Vejalla Jul 5 '19 at 4:59
0
$\begingroup$

If you enjoy harmonic numbers $$S_n=\sum_{i=0}^n \frac 1{n+1}=H_{2 n}-H_{n-1}$$ I you use the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ apply it twice and continue with Taylor series to get $$S_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$ which show the limit and how it is approached.

This also gives a quick approximation of the partial sum. For example $$S_{10}=\frac{178964263}{232792560}\approx 0.7687714$$ while the above expression would give $$\frac{121}{1600}+\log (2)\approx 0.7687722$$

$\endgroup$
0
$\begingroup$

Approximately half of the terms are $\ge\frac{2}{3n}$ and the rest are $\ge\frac1{2n}$ so give or take an error of the order of $1/n$, $S_n$ is at least $$\frac n2\frac2{3n}+\frac n2\frac1{2n}=\frac13+\frac14=\frac{7}{12}.$$ While it's obvious the limit must be at least $1/2$, we have got a lower bound which is strictly greater than that.

$\endgroup$
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{{1 \over n} + \cdots + {1 \over 2n}} = \sum_{k = n}^{2n}{1 \over k} = \sum_{k = 0}^{n}{1 \over k + n} = {1 \over n}\sum_{k = 0}^{n}{1 \over k/n + 1} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,& \int_{0}^{1}{\dd x \over x + 1} = \bbx{\ln\pars{2}} \end{align}

$\endgroup$
  • $\begingroup$ Quote from the question: “I know that it converges to Ln(2) but we still haven't learned about integrals in my class, so calculating its limits kills the purpose of the exercise.” $\endgroup$ – Martin R Jul 5 '19 at 6:48
  • $\begingroup$ @MartinR You're right. I'll try to modify my answer. Thanks. $\endgroup$ – Felix Marin Jul 5 '19 at 6:50
  • $\begingroup$ You are welcome to edit your answer. Just note that a possible duplicate target has already been pointed out. $\endgroup$ – Martin R Jul 5 '19 at 6:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.