I'm having problems with the following limit:
$\lim\limits_{n \to \infty}(e-1)\sum_{k=1}^n \frac{1}{n+k(e-1)} $
It's a task from a taskbook for first year engineering students. Any help is appreciated. Thanks.
I'm having problems with the following limit:
$\lim\limits_{n \to \infty}(e-1)\sum_{k=1}^n \frac{1}{n+k(e-1)} $
It's a task from a taskbook for first year engineering students. Any help is appreciated. Thanks.
Hint: Use the Riemman sum definition of definite integral: $$\int_a^bf(x)dx = \lim\limits_{n\to\infty}\dfrac{1}{n}\sum_{k=0}^nf\left(\frac kn\right).$$ Your challenge now would be to suitably choose $f(x).$
Replacing $e-1$ by $a$, you want $s(a) =\lim\limits_{n \to \infty}a\sum_{k=1}^n \frac{1}{n+ka} $.
Let $t(a, n) =\sum_{k=1}^n \frac{1}{n+ka} $.
Then
$\begin{array}\\ t(a, n) &=\sum_{k=1}^n \frac{1}{n+ka}\\ &=\frac1{n}\sum_{k=1}^n \frac{1}{1+a(k/n)}\\ &\to\int_0^1\frac{dx}{1+ax} \qquad\text{Riemann}\\ &=\frac1{a}\int_0^a\frac{dy}{1+y} \qquad y=ax, dx = dy/a\\ &=\frac1{a}\ln(1+y)|_0^a &=\frac{\ln(1+a)}{a} \end{array} $
so
$s(a) =a\frac{\ln(1+a)}{a} =\ln(1+a) $.
If $a=e-1$, $s(a) =\ln(e) =1 $.
Without Riemman sum assuming that you enjoy harmonic numbers.
Making the problem more general, consider $$a\,S_n=a\sum_{k=1}^n \frac{1}{n+k\,a}=H_{\left(1+\frac{1}{a}\right) n}-H_{\frac{n}{a}}$$
Now, using the asymptotics $$H_p=\gamma +\log (p)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ apply it twice and continue with Taylor expansion to get $$a\, S_n=\log(1+a)-\frac{a^2}{2 (a+1) n}+\frac{a^3(a+2)}{12 (a+1)^2 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.
Now, if you make $a=e-1$ $$(e-1)\,S_n=1-\frac{(e-1)^2}{2 e n}+\frac{(e-1)^3 (1+e)}{12 e^2 n^2}+O\left(\frac{1}{n^4}\right)$$