# Show $S_n = \frac{n}{2n+1}$ converges to $\frac{1}{2}$

This is my thinking so far but it is by no means a rigorous proof:

$\frac{n}{2n+1} < \frac{n}{2n} = \frac{1}{2}$

I don't know how to translate this result into a proof, thank you.

• Showing something is less than something else, certainly doesn't imply it converges to it! n/4n < n/2n but they obviously don't converge. Try instead noting n/2n+1 = 1/2+(1/n) and 1/n converges to 0. Jan 15 '16 at 7:35
• @fleablood While I agree with the first half of your comment, $n/(2n+1)\neq1/2+(1/n)$. Jan 15 '16 at 19:04
• @fleablood Sorry I think it was just a question of parentheses, but I believe the asker is looking for a rigorous proof. Jan 15 '16 at 19:06
• In comments I am much more lazy about LaTex then in answers (because of lack of preview and under the assumption we aren't doing any complicated concepts requiring precise math) and I assumed, perhaps in correctly and inappropriately, that in context it was clear I mean $\frac {1}{2 + \frac 1 n}$. Although looking at my comment, I must admit, it really does look like $\frac 1 2 + \frac 1 n$ and (oh, that converges to 1/2 too in the "idiot coincidence" way!) ... so, I apologize. Jan 15 '16 at 19:32
• Yes, I know the user wanted a rigorous proof and in the comments (not the answers) I gave a hint but not a proof. The main purpose of my comment was to point out that translating $f(n) < g(n)$ therefore $\lim f(n) = \lim g(n)$ ... um... isn't going to work. Jan 15 '16 at 19:35

$\frac{n}{2n+1}=\frac{n}{n(2+\frac{1}{n})}=\frac{1}{2+\frac{1}{n}}\to\frac{1}{2}$

If you want a rigorous proof with all the $\varepsilon$'s then note that

$$S_n=\frac n{2n+1} = 1 - \frac{n+1}{2n+1} > \frac 12 - \frac 1{2n+1} > \frac 12 - \frac1n$$

Now you have that

$$\frac 12 > S_n > \frac 12 - \frac 1n,$$

so

$$\left|S_n - \frac 12\right| < \frac 1n$$

Now by the Archimedean property, for all $\varepsilon > 0$ there exists $N\in\mathbb{N}$ such that for all $n> N$, $\frac 1n < \varepsilon$. Thus $S_n\to \frac 12$.

The expression $\displaystyle\lim_{n\to\infty}a_n=\alpha$ means that there always exists an integer $N$ for every positive number $\epsilon$ such that $\lvert a_n-\alpha\rvert$ is smaller than $\epsilon$ if $n$ is bigger than $N$. In symbolic words,

$$(\forall\epsilon>0)(\exists N\in\mathbf{N})(\forall n\in\mathbf{N})(n>N\implies\lvert a_n-\alpha\rvert<\epsilon).$$

Following this definition, let's see whether $S_n$ converges and indentify what the value is if it does. Let $\epsilon$ be a sufficiently small but positive number and $n>\dfrac{1}{4}\left(\dfrac{1}{\epsilon}-2\right)$. From this inequality, we easily get $$\frac{1}{2}-S_n<\epsilon.$$ Since $S_n$ is always smaller than $\dfrac{1}{2}$ as you mentioned, we get \begin{gather} \frac{1}{2}-S_n<\epsilon\ \land\ S_n-\frac{1}{2}<(0<)\epsilon, \\ \therefore\quad\frac{1}{2}-\epsilon<S_n<\frac{1}{2}+\epsilon. \end{gather}

This finally implies $$\left\lvert S_n-\frac{1}{2}\right\rvert<\epsilon\quad \text{for any}\ n>\frac{1}{4}\left(\frac{1}{\epsilon}-2\right).$$

Thus, $\displaystyle\lim_{n\to\infty}S_n=\dfrac{1}{2}$ (and $N=\left\lfloor\dfrac{1}{4}\left(\dfrac{1}{\epsilon}-2\right)\right\rfloor+1$).