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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.

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  • $\begingroup$ What have you tried? $\endgroup$ – Viktor Glombik Jan 25 at 22:25
  • $\begingroup$ I've tried to simplify the sum (not sure if there is any way to do it after some time of thinking) and I've also tried to use the squeeze theorem. Both failed. $\endgroup$ – Rafał Szypulski Jan 25 at 22:28
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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 $f(x).$

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  • $\begingroup$ I think you mean $\int_0^1f(x)\;dx$. $\endgroup$ – adfriedman Jan 26 at 0:43
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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 $.

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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 $a=e-1$ $$(e-1)\,S_n=1-\frac{(e-1)^2}{2 e n}+\frac{(e-1)^3 (1+e)}{12 e^2 n^2}+O\left(\frac{1}{n^4}\right)$$

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