1

We know,

$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^nf\left(\frac rn\right)=\int_0^1f(x)dx$$

Now,we have $\int_0^1\frac{\log( 1+x)}{x}dx$ So,$f(x)=\frac{\log (1+x)}{x}$.So we find $f(r/n)$ by replacing $x$ with $r/n$. Then putting in the summation form on LHS.

Converting to summation, $$\lim_{n\to\infty}\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}$$

Now this should evaluate to $1-\frac{1}{2^2}+\frac1{3^3}-\cdots$.How do I carry on the simplification?

Thanks for any help!

Batominovski
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Soham
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  • Assuming $log(x)$ here is a natural log ( or it can be converted to one by appropriately scaling it). You can use the taylor series approximation of log(1+x) and further for cases where x tends to 0 ( because x = r/n and n tends to infinity) log(1+x) can be approximated to just x by ignoring the higher power terms in the taylor series. – Aditya Sep 02 '18 at 08:06
  • https://math.stackexchange.com/questions/1093132/easiest-way-to-calculate-int-01-frac-log-1xx-dx , https://math.stackexchange.com/questions/2029886/evaluate-int-01-frac-log1xx?noredirect=1 , https://math.stackexchange.com/questions/1338307/hint-to-integrate-int-01-frac-log1xx-mathrmdx – V.G Jan 28 '21 at 17:39
  • https://math.stackexchange.com/questions/288830/finding-int1-0-frac-log1xxdx-without-series-expansion – V.G Jan 28 '21 at 17:40

2 Answers2

3

HINT:

Write \begin{align}\frac1r\log\left(1+\frac rn\right)&=\frac1r\left[\frac rn-\frac12\left(\frac rn\right)^2+\frac13\left(\frac rn\right)^3-\cdots\right]\\&=\frac1n-\frac r{2n^2}+\frac{r^2}{3n^3}-\cdots\end{align} so \begin{align}\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}&=1-\frac{\sum r}{2n^2}+\frac{\sum r^2}{3n^3}-\cdots\\&=1-\frac12\int_0^1r\,dr+\frac13\int_0^1r^2\,dr-\cdots\end{align} as each term is essentially a Riemann sum.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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    Or treat each term $$\frac{(-1)^k}{(k+1),n^{k+1}},\sum_{r=1}^n,r^k$$ as the Riemann sum for $$\frac{(-1)^k}{(k+1)},\int_0^1,x^k,\text{d}x,.$$ – Batominovski Sep 02 '18 at 08:09
  • @Batominovski Nice. I'll add this in. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 02 '18 at 08:10
  • Can't this be done without using $\log(1+x)$ expansion? I was trying $\lim_{n\to\infty}\sum_{r=1}^n\frac{\log(1+\frac rn)}{r}=\lim_{x\to0}\sum_{r=1}^n\frac{\log(1+x)}{x}\cdot\frac1n$ where $x=\frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here? – Soham Sep 02 '18 at 08:13
  • @tatan You can't take limits whilst the term is being multiplied by another in a summation. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 02 '18 at 08:15
  • @TheSimpliFire Okay thanks. But why can you please explain? – Soham Sep 02 '18 at 08:15
  • @tatan because otherwise you'll end up with $$\sum_{r=1}^n\frac1n=1$$ which is absurd since the integral is less than $1$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 02 '18 at 08:20
  • It can be seen that $$I:=\int_0^1,\frac{\ln(1+x)}{x},\text{d}x=\int_0^\infty,\ln\big(1+\exp(-t)\big),\text{d}t,,$$ where $t:=-\ln(x)$. Using integration by parts, we get $$I=\int_0^\infty,\frac{t}{\exp(t)+1},\text{d}t,.$$ As $\eta(s)=\displaystyle\frac1{\Gamma(s)}\int_0^\infty,\frac{u^{s-1}}{\exp(u)+1},\text{d}u$ for all complex number $s$ with $\text{Re}(s)>0$, where $\Gamma$ is the usual gamma function and $\eta$ is the Direchlet eta function, we obtain $$I=\eta(2)=\sum_{r=1}^\infty,\frac{(-1)^{r-1}}{r^2}=\frac{\pi^2}{12},.$$ – Batominovski Sep 02 '18 at 08:33
  • If you don't want to expand $\ln(1+x)$, then you can expand $$\frac{t}{\exp(t)+1}=\sum_{r=1}^\infty,(-1)^{r-1},t,\exp(-rt)$$ instead. You can also use complex analysis to prove that $I=\eta(2)=\dfrac{\pi^2}{12}$ without using any series expansion. – Batominovski Sep 02 '18 at 08:35
  • @TheSimpliFire Oopps. I meant to put the two comments above under the OP's question. Wrong place. But well, they are here already. So I am leaving them here. I hope you don't mind. I didn't write these comments as a separate answer since the OP clearly wanted to use Riemann sums, but my idea doesn't use any Riemann sums. – Batominovski Sep 02 '18 at 08:41
1

HINT

Following the previous suggestion by TheSimpliFire

$$\sum_{r=1}^n\frac{\log\left(1+\frac{r}{n}\right)}{r}=1-\frac{\sum r}{2n^2}+\frac{\sum r^2}{3n^3}-\cdots\to1-\frac1{2\cdot 2}+\frac1{3\cdot 3}\ldots=$$$$=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^2}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12}$$

indeed from the well known results

we have

$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}=\sum_{k=1}^\infty \frac{1}{(2k-1)^2}-\frac14\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}8-\frac{\pi^2}{24}=\frac{\pi^2}{12}$$

user
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  • Can't this be done without using $\log(1+x)$ expansion? I was trying $\lim_{n\to\infty}\sum_{r=1}^n\frac{\log(1+\frac rn)}{r}=\lim_{x\to0}\sum_{r=1}^n\frac{\log(1+x)}{x}\cdot\frac1n$ where $x=\frac rn$... Now the log part goes to $1$ (by using standard limit formula). Can the answer be derived from here? – Soham Sep 02 '18 at 08:15
  • @tatan No we can't since we need all the terms of the expansion and take 1 corresponds to take only one terms which leads to a wrong result. – user Sep 02 '18 at 08:31