Evaluate $\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n})$

Question

Show that $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6$$ Here I need to use the definition of integral but I faced problem in range . Please help.

Answer 1

Maybe it is intended that you mention Riemann sums explicitly. Rewrite our sum as $$\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \frac{1}{1+\frac{3}{n}}+\cdots +\frac{1}{1+\frac{5n}{n}} \right).$$ We recognize this as a (right) Riemann sum for the integral $$\int_0^5 \frac{dx}{1+x},$$ which has value $\log 6$.

For imagine dividing the interval from $x=0$ to $x=5$ into equal subintervals of width $\frac{1}{n}$. The right ends of these subintervals are at $\frac{1}{n}$, $\frac{2}{n}$, $\frac{3}{n}$, and so on up to $\frac{5n}{n}$. So if $f(x)=\frac{1}{1+x}$, then our summands are precisely the function values at these right endpoints. The limit of the Riemann sum as $n \to\infty$ is therefore $\log 6$.

Answer 2

$$\int_{n+k}^{n+k+1}\frac{\mathrm dt}t\leqslant\frac1{n+k}\leqslant\int_{n+k-1}^{n+k}\frac{\mathrm dt}t$$ $$ \log\left(6-\frac5{n+1}\right)=\int_{n+1}^{6n+1}\frac{\mathrm dt}t\leqslant\sum_{k=1}^{5n}\frac1{n+k}\leqslant\int_{n}^{6n}\frac{\mathrm dt}t=\log6$$

Answer 3

The answers already provided say it all but i will explain in detail the one by André Nicolas so that you can understand how it follows from the definition of integral.

Derivation of the relation we'll use

We know that by definition (that fancy figure you find in all textbooks to give geometrical interpretation of integral) the integral can be viewed as the sum of little areas under a curve within an interval.

Thus let $f(x)$ be a single valued continuous function defined in (a,b) such that $b \gt a$ and let (a,b) be divisible into $n$ equal parts each of length $\Delta x$ such that $n\Delta x=b-a$ then

$$ \int \limits_{a}^{b} {f(x) dx} = \lim_{n \to \infty} \Delta x[f(a)+f(a+\Delta x)+f(a+2\Delta x)+\cdots+f(a+(n-1)\Delta x)] \\ = \lim_{n \to \infty} \Delta x\sum_{i=0}^{n-1} f(a+i\Delta x) $$

Now we know that $n\Delta x=b-a$. Let $a=0$ and $b=1$,
so we have $\Delta x=\frac{1}{n}$
And finally

$$ \lim_{n \to \infty} \frac{i}{n}\sum_{i=0}^{n-1} f(\frac{i}{n}) = \int \limits_{0}^{1} {f(x) dx} $$

This last integral is the one that will allow you to compute limit of series using integration. But of course it comes with a condition:
The series must have $\frac{1}{n}$ as a common factor such that it tends to zero as $n \mapsto 0$ and the function $f(i/n)$ must give back the original series under the summation sign and a multiplication by $\frac{1}{n}$.

Computing the required limit

In order to find the limit of our series using the above relation here is a path we'll follow to find the answer:

1. Find the general term of the series,
2. Take out $\frac{1}{n}$ from the general term such that you find something similar to the relation above, you will have another $\frac{1}{n}$ inside the function,
3. Do the calculation.

And here we go!!

1. Let's find the general term:
   The series is given by $(\frac{1}{n+}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{6n})$ which we rewrite as $(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+5n})$ .
   The general term is given by $\frac{1}{n+i}$ such that the sum is defined from $i=1$ up to $i=5n$.

2. Taking $\frac{1}{n}$ out:
   We get $$ \frac{1}{n}\left(\frac{1}{1+\left(\frac{i}{n}\right)}\right) $$

3. The calculation:
   From the general term, we have the sum:
   $$ \sum_{i=1}^{5n} \frac{1}{n}\left(\frac{1}{1+\left(\frac{i}{n}\right)}\right) $$

But this is unusable with regard to the relation found above, so let's make one transformation.
Let $k=5n$, then $n=\frac{k}{5}$.
Replacing in the sum above we get:
$$ \sum_{i=1}^{k} 5\frac{1}{k}\left(\frac{1}{1+5\left(\frac{i}{k}\right)}\right) $$ Which now is something we can use!
We get:
$$ \int \limits_{0}^{1} {5\left(\frac{1}{1+5x}\right) dx} $$ We do a change of variable and we have such that for $t=5x$ we get $t=0$ for $x=0$, $t=5$ for $x=1$ and $dt=5dx$.
And last we get the integral provided by André Nicolas:
$$ \int \limits_{0}^{5} {\frac{1}{1+t} dt} $$ Which is simply $\log(6)$.

Hope that was helpful and correct!

Answer 4

If you can use the theorem: (1)

$$\lim_{n\to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n\right)=c$$

(the constant c is the Euler-Mascheroni constant $\gamma$ but it's irrelevant here)

Then this is also true:

$$\lim_{6n\to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{6n}-\log 6n\right)=c$$

But when $6n\to \infty$, then $n\to \infty$, too. So we have: (2)

$$\lim_{n\to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{6n}-\log 6n\right)=c $$

We also have: (3)

$$\log 6n - \log n = \log 6$$

Combining (1), (2) and (3) gives the wanted limit:

$$(2)-(1)+(3) = c - c + \log 6 = \log 6$$

Answer 5

Another way to do this is to rewrite the limit in the form $$ \lim_{n\to\infty}\sum_{j=1}^{5n}\frac{1}{1+\frac{j}{n}}\,\,\, \frac{1}{n} $$ and notice that the sum is a Riemann sum for $\int_1^6 \frac{1}{t}\,dt$ with $5n$ subdivisions (so that $\Delta t=\frac{1}{n}$).

Answer 6

If $H_x = 1+\frac{1}{2}+\cdots+\frac{1}{x}$ (harmonic numbers) your sum is equal to

$$\lim_{n \to \infty} H_{6n}-H_n$$

Recall that $H_x=\log(n)+\gamma+O\left(\frac{1}{n}\right)$, so we see

$$\lim_{n \to \infty} H_{6n}-H_n= \lim_{n \to \infty}(\log (6n)+\gamma)-(\log (n)+\gamma)+O\left(\frac{1}{n}\right)= \lim_{n \to \infty}\log 6+O\left(\frac{1}{n}\right)=\log 6$$