Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$

Question

$$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$ the task is to find $$\lim_{n \to \infty} S_n$$ I've tried different ways, but all I could do is to make an estimation that the limit is somewhere between 0.5 and 1, but that's not the exact answer.

Manually doing fist elements gives

$$\frac{1}{3}, \quad \frac{1}{5} + \frac{1}{7}, \quad \frac{1}{7} + \frac{1}{9} + \frac{1}{11}, \quad ...$$

I'm trying to restate it as $$\lim_{n \to \infty} \bigg[ \sum_{i = 1}^{\infty} \frac{1}{2i - 1} - \sum_{i = 1}^{n} \frac{1}{2i - 1} - \sum_{i = 4n}^{\infty} \frac{1}{2i - 1} \bigg]$$

so that I cut off the beginning of the series and it's tail.

Answer 1

I'll give a hint that follows user43208's reasoning (though NB that the formula he gives is missing a numerical factor from one term and so leads to an incorrect result).

Hint The quantity $S_n$ can be written as $$\left(\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + \cdots + \frac{1}{4 n - 1} + \frac{1}{4n}\right) \\ \quad- \left(\frac{1}{2n + 2} + \frac{1}{2 n + 4} + \cdots + \frac{1}{4 n - 2} + \frac{1}{4 n} \right),$$ and the second quantity in parentheses can be written as $$\frac{1}{2}\left(\frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2 n - 1} + \frac{1}{2 n}\right).$$

So we can write $$S_n = (H_{4n} - H_{2n}) - \tfrac{1}{2}(H_{2n} - H_n),$$ where $H_m$ is the $m$th harmonic number, $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{m}$. On the other hand, $H_n$ satisfies the asymptotic expansion $$H_m = \log m + \gamma + O\left(\frac{1}{m}\right) .$$ Here, $\gamma$ is the Euler-Mascheroni constant---in fact, it is usually essentially defined by this previous equation, but substituting the expansion in our above formula for $S_n$ shows that its value is irrelevant for purposes of evaluating our limit.

If the question had instead by tagged $\texttt{calculus}$, the intended method would presumably be to recognize $S_n$ as a Riemann sum, as Fedor Petrov's hint does. (NB the answer he states is also off by a numerical factor.)

Answer 2

Consider the following inequalities:

$$ \begin{align} \frac{1}{2n+1+2x} &\le \frac{1}{2n+1} \le \frac{1}{2n-1+2x}, \quad 0\le x\le 1, \quad (1) \\ \frac{1}{2n+1+2x} &\le \frac{1}{2n+3} \le \frac{1}{2n-1+2x}, \quad 1\le x\le 2 , \quad (2)\\ \frac{1}{2n+1+2x} &\le \frac{1}{2n+5} \le \frac{1}{2n-1+2x}, \quad 2\le x\le 3 , \quad (3)\\ \frac{1}{2n+1+2x} &\le \frac{1}{2n+7} \le \frac{1}{2n-1+2x}, \quad 3\le x\le 4 , \quad (4)\\ ... \quad & \le \quad ... \quad \le \quad... \\ \frac{1}{2n+1+2x} &\le \frac{1}{2n+2n-1} \le \frac{1}{2n-1+2x}, \quad n-1\le x\le n. \quad (5) \end{align} $$

Integrating each inequality over the corresponding interval, side by side. For example:

$$ \begin{align} \int_0^1 (1) dx \Rightarrow \int_0^1 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+1} \le \int_0^1 \frac{dx}{2n-1+2x}, \quad (1') \\ \int_1^2 (2) dx \Rightarrow \int_1^2 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+3} \le \int_1^2 \frac{dx}{2n-1+2x}, \quad (2') \\ \int_2^3 (3) dx \Rightarrow \int_2^3 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+5} \le \int_2^3 \frac{dx}{2n-1+2x}, \quad (3') \\ ... \quad & \le \quad ... \quad \le \quad... \\ \int_{n-1}^n (5) dx \Rightarrow \int_{n-1}^n \frac{dx}{2n+1+2x} &\le \frac{1}{2n+2n-1} \le \int_{n-1}^n \frac{dx}{2n-1+2x}. \quad (5') \end{align} $$ Summing $(1')$ through $(5')$, side by side:

$$ \begin{align}\int_0^n \frac{dx}{2n+1+2x} &\le S_n \le \int_0^n \frac{dx}{2n-1+2x} \quad (6)\\ \frac12 \ln|2n+1+2x|_0^n &\le S_n \le \frac12 \ln |2n-1+2x|_0^n \\ \frac12 \ln\left|\frac{4n+1}{2n+1}\right| &\le S_n \le \frac12 \ln \left|\frac{4n-1}{2n-1}\right|, \quad (7) \end{align}$$

Thus, from $(7)$:

$$ \lim_{n\to\infty} S_n = \frac{\ln 2}{2} $$

Answer 3

$\displaystyle \sum\limits_{k=2n+1}^{4n-1} \frac{1}{k} - \sum\limits_{k=n+1}^{2n-1} \frac{1}{2k} \approx (\ln 4 – \ln 2) - \frac{1}{2}(\ln 2 – \ln 1) = \frac{\ln 2}{2} $

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Explanation (with suitable variables $a,b,n,q,u,v$, e.g. $a\in\mathbb{N}$, $n\to\infty$):

$\displaystyle \prod\limits_{k=1}^{an} (1+\frac{x}{k}) \approx \frac{(an)^x}{\Gamma(1+x)}$

$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an} (1+\frac{x}{k}) \approx (\frac{(an)^x}{\Gamma(1+x)})/(\frac{n^x}{\Gamma(1+x)}) = a^x $

$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an} (1+\frac{x}{qk})^q \approx a^x $ , $\hspace{2cm} q\to\infty$ :

$=>\hspace{1cm}\displaystyle \prod\limits_{k=n}^{an} \exp(\frac{x}{k}) \approx a^x$ , $\hspace{2cm} log$ :

$=> \hspace{1cm}\displaystyle \sum\limits_{k=n}^{an} \frac{1}{k} \approx \ln a $

$=> \hspace{1cm}\displaystyle \sum\limits_{k=bn+u}^{an+v} \frac{1}{k} \approx \ln a - \ln b \enspace\enspace$ with constants $\enspace|u|,|v|\ll n $

Answer 4

The sum can be written as

$$S(n) = \sum_{k=n}^{2n} \frac{1}{2k-1}$$

If we replace the sum by an integral we find

$$I(n) = \int_n^{2 n} \frac{1}{2 k-1} \, dk = \frac{1}{2} \log (2+\frac{1}{2n-1})$$

And the limit $n\to\infty$ is

$$S(n\to\infty) = \frac{1}{2} \log (2)\tag{1}$$

The justification of the replacement of the sum by an integral is easily done considering the inequalities (lower "sum" < S < upper "sum")

$$\int_n^{2 n} \frac{1}{2 k-1} \, dk < S(n) < \int_n^{2 n} \frac{1}{2 (k-1)-1} \, dk $$

which are obvious graphically.

Hence we have

$$ \frac{1}{2} \log \left(2+\frac{1}{2n-1}\right)< S(n) < \frac{1}{2} \log \left(2 + \frac{3}{2 n-3}\right)$$

and the limit $n\to\infty$ gives back the result (1).

Generalization

It is tempting to consider the more general sum (where $0<a<b, c<0, d<a \;c \;n$)

$$S(a,b,c,d,n)=\sum _{k=a\; n}^{b\; n} \frac{1}{c\; k-d}$$

The limit is similarly calculated with the result

$$\lim_{n\to \infty } \, S(a,b,c,d,n)=\frac{1}{c}\;\log \left(\frac{b}{a}\right)$$

Notice the independence of $d$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n}{1 \over 2n + 2k - 1} & = {1 \over 2}\sum_{k = 1}^{\infty} \pars{{1 \over k + n - 1/2} - {1 \over k + 2n - 1/2}} = {H_{2n - 1/2}-H_{n - 1/2} \over 2} \\[5mm] & \stackrel{\large\mrm{as}\ n\ \to\ \infty}{\large\sim}\,\,\, {\ln\pars{2n + 1/2} - \ln\pars{n + 1/2} \over 2} \,\,\,\stackrel{\large\mrm{as}\ n\ \to\ \infty}{\large\to}\,\,\, \bbx{\ds{{1 \over 2}\,\ln\pars{2}}} \end{align}