Make me an integral

Calculus Level 3

$\lim_{n\to\infty}\left(\frac{1}{a+n}+\frac{1}{2a+n}+\frac{1}{3a+n}+\ldots+\frac{1}{na+n}\right)$

Find the "closed form result" of the limit above.

Here, $a\gt 0$ is an arbitrary constant such that the expression inside the limit is defined.

\pi^e\cdot \ln(a)

\pi^a

\dfrac{\ln(\sqrt{a}+1)}{a}

\dfrac{\ln(a+1)}{a}

\dfrac{\ln(a+2)}{a}

e^a

a

\dfrac{\log_{10}(a+1)}{a}

2 solutions

Akiva Weinberger
May 9, 2015

But I don't want to make you an integral!

Substitute $ax$ for $n$ .

$\displaystyle\phantom{=\frac1a}\lim_{x\to\infty}\left(\frac1{a+ax}+\frac1{2a+ax}+\dotsb+\frac1{a^2x+ax}\right)$

$\displaystyle=\frac1a\lim_{x\to\infty}\left(\frac1{1+x}+\frac1{2+x}+\dotsb+\frac1{ax+x}\right)$

$\displaystyle=\frac1a\lim_{x\to\infty}\left(H_{ax+x}-H_x\right)\quad$ where $H_x=1+\dotsb+\frac1x$

Since $H_x\approx \ln x+\gamma$ as $x$ grows larger, we have:

$\displaystyle=\frac1a\lim_{x\to\infty}\big((\ln(ax+x)-\gamma)-(\ln(x)-\gamma)\big)=\frac1a\lim_{x\to\infty}\ln\left(\frac{ax+x}x\right)$

$\displaystyle=\frac1a\lim_{x\to\infty}\ln(a+1)=\boxed{\dfrac{\ln(a+1)}a}$

Tanishq Varshney
Apr 18, 2015

$\displaystyle \lim_{n\to \infty} \sum^{n}_{r=1}\frac{1}{ar+n}$

$\displaystyle \lim_{n\to \infty} \sum^{n}_{r=1} \frac{1}{n}(\frac{1}{a\frac{r}{n}+1})$

$\displaystyle \int^{1}_{0} \frac{1}{ax+1}.dx$

$\frac{1}{a} ln(ax+1) |^{1}_{0}$

After applying limits

$\frac{ln(a+1)}{a}$

What is the name of this method ? @Tanishq Varshney

Refaat M. Sayed - 6 years, 1 month ago

This method actually uses the Riemann sum form of a definite integral. We have, for a Riemann-integrable function $f(x)$ on $[0,b]$ ,

$\large\lim_{n\to\infty}\sum_{k=1}^{bn}f\left(\frac{k}{n}\right)=\int\limits_0^b f(x)\,\mathrm dx$

where $b\in\Bbb{Z^+}$ and $n=\dfrac{b}{h}~,~h\to 0$

Prasun Biswas - 6 years, 1 month ago