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Struggling to solve this problem,

$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$

My approach:

$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +...+ \frac{1}{6n}\right)$

= $\displaystyle\lim\limits_{n \to \infty} \int_{0}^{1}(x^{n-1} + x^{n} + x^{n-2}+...+ x^{6n-1}) dx$

= $\displaystyle\lim\limits_{n \to \infty}\int_{0}^{1}\left(x^{n-1} \cdot \frac{x^{5n+1} - 1}{x-1}\right)dx$

got stuck here and don't know how to solve it further (other approaches which are simpler would also help)

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    $\begingroup$ Should there be $x$ in the limit? Also what about the limits of integration? $\endgroup$
    – mowzorn
    Jan 6, 2023 at 10:30
  • $\begingroup$ @SineoftheTime I have corrected it now, sorry! $\endgroup$ Jan 6, 2023 at 10:49
  • $\begingroup$ @ArunMadhav no need to be sorry, I'll delete my comment ;) $\endgroup$ Jan 6, 2023 at 10:50
  • $\begingroup$ What is $\lim\limits_{n \to \infty}\frac{1}{n}$, $\lim\limits_{n \to \infty}\frac{1}{n+1}$, ...? From here you can calculate the limit of the sum. $\endgroup$
    – bayes2021
    Jan 6, 2023 at 10:51
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    $\begingroup$ No, because in each 'step' the number of terms in the sum changes. $\endgroup$
    – mowzorn
    Jan 6, 2023 at 10:54

4 Answers 4

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You also can use Riemann-sums: $$ \left(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots + \frac{1}{6n}\right) $$ $$ = \left(1+\frac{n}{n+1}+\frac{n}{n+2}+\dots + \frac{n}{n+ (5n-1)}\right)\frac{1}{n}+ \frac{1}{6n} $$ $$ =\left(1+\frac{1}{1+1/n}+\frac{1}{1+2/n}+\dots + \frac{1}{1+(5n-1)/n}\right)\frac{1}{n} + \frac{1}{6n} $$ $$ \to \int_0^5\frac{1}{1+x}dx + 0 = \ln(6) \quad (n \to \infty). $$

Edit: On the interval $[0,5]$ we consider the partition $$ x_k=\frac{k}{n} \quad (k=0, \dots, 5n), $$ and the function $f(x)=1/(1+x)$. Choosing the left boundary point on each interval $[x_k,x_{k+1}]$ we get the Riemann-sum $$ \sum_{k=0}^{5n-1} f(x_k) (x_{k+1}-x_k) = \sum_{k=0}^{5n-1} \frac{1}{1+k/n} \cdot \frac{1}{n}. $$ Now it is known that $$ \sum_{k=0}^{5n-1} f(x_k) (x_{k+1}-x_k) \to \int_0^5 f(x) dx \quad (n \to \infty); $$ see https://en.wikipedia.org/wiki/Riemann_sum, for example.

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  • $\begingroup$ your approach is similar to the solution provided in which I was unable to understand how they took the decision to go from summation(your 3rd step) to integration(4th step), could you explain what is the intuition for converting a summation into integration in general? $\endgroup$ Jan 6, 2023 at 11:48
  • $\begingroup$ I will edit the answer. $\endgroup$
    – Gerd
    Jan 6, 2023 at 11:52
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I will use the fact that the sequence $$x_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}-\ln n$$ has a limit $\gamma = 0.577...$ (known as Euler–Mascheroni constant). Let's denote $$a_n = \frac{1}{n}+\ldots+\frac{1}{6n}.$$ We can see that $$x_{6n}-x_{n-1}=a_n-\ln(6n)+\ln(n-1) = a_n -\ln\left(\frac{6n}{n-1}\right).$$ Then $$\lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x_{6n}-x_{n-1}+\ln\left(\frac{6n}{n-1}\right)\right) = \gamma-\gamma+\ln6 = \ln 6.$$

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Let define $\displaystyle H_n=1+\frac {1}{2}+ \frac {1}{3}...+\frac {1}{n} $ more precisely is the sum up to n $H_n=\sum_{k=0}^n=\frac {1}{k}$.

Using Euler-Maclaurin formula $\displaystyle H_n=ln(n)+\gamma+\frac{1}{2n}-\epsilon_n$ where $\gamma\approx0.5772...$and $0\le\epsilon_n\le\frac{1}{8n^2}$

$\displaystyle H_{6n}=ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}$

Noting that your expression is equal to:

$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$=

$\displaystyle \lim\limits_{n \to \infty} H_{6n}-H_n$ = $\lim\limits_{n \to \infty}$ ($ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-(ln(n)+\gamma+\frac{1}{2n}-\epsilon_n)$)=

$\lim\limits_{n \to \infty}$ ($ln(6)+ln(n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-ln(n)-\gamma-\frac{1}{2n}+\epsilon_n$)= $ln(6)$.

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This method uses neither the Euler-Mascheroni constant, nor Riemann sums. Rather, it uses bounds from the graph of $\ f(x) = \frac{1}{x},\ $ using the fact that $\ f\ $ is decreasing towards $\ 0.$

Whenever $\ k\geq i\geq 1,\ $ we have:

$$\int_{x=i+1}^{x=k+1} \frac{1}{x} dx \leq \sum_{j=i+1}^{j=k} \frac{1}{j} \leq \int_{x=i}^{x=k} \frac{1}{x} dx $$

$$ \implies \ln(k+1) - \ln(i+1) \leq \sum_{j=i+1}^{j=k} \frac{1}{j} \leq \ln(k) - \ln(i). \qquad (1) $$

If $\ n\ $ is large enough, we can substitute $\ k=6n\ $ and $\ k=n-1\ $ into $\ (1)\ $ giving, respectively:

$$ \ln(6n+1) - \ln(i+1) \leq \sum_{j=i+1}^{j=6n} \frac{1}{j} \leq \ln(6n) - \ln(i) \qquad (2) $$

and

$$ \ln(n) - \ln(i+1) \leq \sum_{j=i+1}^{j=n-1} \frac{1}{j} \leq \ln(n-1) - \ln(i). \qquad (3) $$

$(2),\ (3)\implies$

$$\ln\left(\frac{6n+1}{n-1}\right) + \ln\left(\frac{i}{i+1}\right) \leq \sum_{j=n}^{j=6n} \frac{1}{j} \leq \ln\left(\frac{6n}{n}\right) + \ln\left(\frac{i+1}{i}\right). \qquad (4) $$

Taking limits in $\ (4),\ $ first as $\ n\to\infty,\ $ then as $\ i\to\infty,\ $ we get:

$$\ln 6 \leq \lim_{n\to\infty}\sum_{j=n}^{j=6n} \frac{1}{j} \leq \ln6. $$

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  • $\begingroup$ I think in the lower bound of summation the integral in equation (1) should be $i+2$ to $k+1$ or any other value to the right of $i+1$ ? $\endgroup$ Jan 6, 2023 at 14:55
  • $\begingroup$ @ArunMadhav I'm confident that $(1)$ is correct: it uses the fact that $f(x)=1/x$ is decreasing towards $0.$ I've drawn a a sketch and I don't see anything wrong with my sketch, so can you explain what you think is wrong about $(1)$ (or the integral inequality above $(1)$)? $\endgroup$ Jan 6, 2023 at 15:27
  • $\begingroup$ as our summation is from $i+1$ to $k$ and as $f(x)$ is a decreasing function but always greater than $0$ so $i+1$ to $k+1$ would be greater than the summation from $i+1$ to $k$ as we are covering more area...that's why I got that doubt, am I making a mistake? if the summation was from $i$ to $k$ then I totally agree that the integral from $i+1$ to $k$ will be a lower bound. $\endgroup$ Jan 6, 2023 at 15:55
  • $\begingroup$ You say: "(the integral from) $i+1$ to $k+1$ would be greater than the sum from $i+1$ to $k$", but this is not true. As an example, let's look at the area under the graph $\ f(x) = \frac{1}{x}\ $ between $\ x=1\ $ and $\ x=2.\ $ Using the fact that $\ \frac{1}{x} \leq \frac{1}{1}\ \forall x\in [1,2],\ $ it follows that $ \int_{x=1}^{x=2}\frac{1}{x} dx \leq 1\cdot\left(\frac{1}{1}\right) \leq \sum_{j=1}^{j=1} \frac{1}{j}.\ $ Notice that just because the difference between the integration limits is $\ 1\ $ and the difference between the summation limits is zero does not mean the... $\endgroup$ Jan 6, 2023 at 16:46
  • $\begingroup$ ... summation is less than the integral. Your misconception is thinking that $\ \sum_{a}^{b} g(z)\ $ has $\ (b-a)\ $ terms, but actually, it has $\ (b-a+1)\ $ terms. I'll try to edit my answer if I get time later tonight to include a diagram for integral inequality $(1)$ which hopefully better illuminates where the inequality comes from. $\endgroup$ Jan 6, 2023 at 16:46

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