Infinity minus infinity

(To the author: The standard notation for those is $H_n$ . They are called harmonic numbers .)

In any case, we have $S_{2n}-S_n=1-\frac12+\frac13-\dotsb-\frac1{2n}$ . Proof:

Proof

Letting $n$ go to infinity, we get the familiar infinite series $1-\frac12+\dotsb=\ln2$ .

Recall $S_{N} \sim \ln N$ . Hence, we have

$\lim_{n \rightarrow \infty} S_{2n}-S_{n} = \lim_{n \rightarrow \infty} \ln(\frac{2n}{n}) = \boxed{\ln 2}$

${ S }_{ 2n }-{ S }_{ n }$ $=\sum _{ r=1 }^{ n }{ \left( \frac { 1 }{ 2n-1 } +\frac { 1 }{ 2n } -\frac { 1 }{ n } \right) }$ $=\sum _{ r=1 }^{ n }{ \left( \frac { 1 }{ 2n-1 } -\frac { 1 }{ 2n } \right) }$

$\lim _{ n\rightarrow \infty }{ (\sum _{ r=1 }^{ n }{ \left( \frac { 1 }{ 2n-1 } -\frac { 1 }{ 2n } \right) } ) }$ = $\log _{ e }{ 2 }$

We see that $S_{n}= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ $...... (A)$ Similarly, $S_{2n}= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}$ $......(B)$ Subtract the above two to get: $S_{2n} - S_{n} = \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} ... + \frac{1}{2n}$ We need to find: $\lim_{n\rightarrow \infty} S_{2n} - S_{n}$ The problem can now be neatly written as: $S= \lim_{n\rightarrow \infty} \sum_{r=1}^n \frac{1}{n+r}$ Or, $S= \lim_{n\rightarrow \infty}\frac{1}{n} \sum_{r=1}^n \frac{1}{1+\frac{r}{n}}$ Now make the substitution- $\frac{r}{n} \rightarrow x$ and $\frac{1}{n} \rightarrow dx$ So now we have $S= \int_{0}^1 \frac{1}{1+x} dx$ Or, $S= \ln(2) = \boxed{0.693}$

I'll add some background and comments to the solution.

The sum $S_n$ is usually known as the $n$ -th harmonic number, denoted by $H_n$ . The asymptotic behavior of $H_n$ is well-known, which follows from the fact that $H_n\sim\int_1^n\frac1xdx=\log(x),$ where $\log(x)$ is the napierian logarithm. In fact, the limit $\lim_{n\to\infty}(H_n-\log(n))$ exists and equals $\gamma$ , the Euler-Mascheroni constant . The fact that the sequence $H_n-\log(n)$ is convergent can be easily shown, first proving that it is bounded, and then proving that it is decreasing.

Therefore, $\lim_{n\to\infty}(S_{2n}-S_n)=\lim_{n\to\infty}(\gamma+\log(2n)-\gamma-\log(n))=\log(2).$ Moreover, we can also notice that in general, if $k$ is a positive integer, $\lim_{n\to\infty}(S_{kn}-S_n)=\lim_{n\to\infty}(\gamma+\log(kn)-\gamma-\log(n))=\log(k).$ As a matter of fact, this problem admits an easy generalization. Let $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ two sequences of positive integers such that diverge to $\infty$ , and $\lim_{n\to\infty}\frac{a_n}{b_n}=L$ exists. Then $\lim_{n\to\infty}(S_{a_n}-S_{b_n})=\log(L).$ As a curious example, $\lim_{n\to\infty} (S_{\lfloor n! e^n\rfloor}-S_{\lfloor n^n \sqrt{n}\rfloor})=\log(\sqrt{2\pi}),$ where $\lfloor x\rfloor$ is the floor function. This follows from Stirling's approximation .

Let $B_n = S_{2n} - S_n = \sum_{x = n + 1}^{2n} \frac{1}{x}$ . We will determine the value of $\lim_{n \to \infty} B_n$ using the Squeeze Theorem (for sequences). We also define $A_n, C_n$ like this: $\begin{aligned} A_n &= \int_{x = n + 1}^{2n + 1} \frac{1}{x} \mathrm{dx} \\ C_n &= \int_{x = n}^{2n} \frac{1}{x} \mathrm{dx}. \end{aligned}$ Since $\frac{1}{x}$ is strictly decreasing for $x > 0$ , by evaluating the integral for $A_n$ using a left Riemann sum with $n$ intervals, we get the following inequality: $\int_{x = n + 1}^{2n + 1} \frac{1}{x} \mathrm{dx} \leq \sum_{x = n + 1}^{2n} \frac{1}{x}.$ Similarly, using a right Riemann sum with $n$ intervals to evaluate the integral for $C_n$ we get the following inequality: $\int_{x = n}^{2n} \frac{1}{x} \mathrm{dx} \geq \sum_{x = n + 1}^{2n} \frac{1}{x}.$ Putting these together we get that $A_n \leq B_n \leq C_n$ . But we can also evaluate $A_n, C_n$ directly by integrating, giving us: $\begin{aligned} A_n &= \log(2n + 1) - \log(n + 1) \\ &= \log\left(\frac{2n + 1}{n + 1}\right) \\ &= \log\left(2 - \frac{1}{n + 1}\right) \\ C_n &= \log(2n) - \log(n) \\ &= \log 2. \end{aligned}$ Therefore, we have the following inequality: $\log\left(2 - \frac{1}{n + 1}\right) \leq S_{2n} - S_n \leq \log 2.$ Therefore, by the Squeeze Theorem we have: $\lim_{n \to \infty} S_{2n} - S_n = \boxed{\log 2}.$