# Limits problem (unable to solve further)

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)

• Should there be $x$ in the limit? Also what about the limits of integration? Jan 6, 2023 at 10:30
• @SineoftheTime I have corrected it now, sorry! Jan 6, 2023 at 10:49
• @ArunMadhav no need to be sorry, I'll delete my comment ;) Jan 6, 2023 at 10:50
• 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. Jan 6, 2023 at 10:51
• No, because in each 'step' the number of terms in the sum changes. Jan 6, 2023 at 10:54

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.

• 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? Jan 6, 2023 at 11:48
• I will edit the answer.
– Gerd
Jan 6, 2023 at 11:52

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 $$\lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x_{6n}-x_{n-1}+\ln\left(\frac{6n}{n-1}\right)\right) = \gamma-\gamma+\ln6 = \ln 6.$$

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:

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

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

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

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:

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

• 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 $i+1$ ? Jan 6, 2023 at 14:55
• @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 $(1)$)? Jan 6, 2023 at 15:27
• 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. Jan 6, 2023 at 15:55
• 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... Jan 6, 2023 at 16:46
• ... 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. Jan 6, 2023 at 16:46