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$\underset{n \to \infty}{\lim} \dfrac{1}{n} \bigg( \dfrac{n}{n+1} + \dfrac{n}{n+2} + \cdots + \dfrac{n}{2n} \bigg)$ is equal to

1. $\infty$
2. $0$
3. $\log_e 2$
4. $1$
in Calculus
recategorized | 120 views
0
Is the answer $0$?
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Nope. It's definitely not $0$.
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Right.
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Why not limit value  zero
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Just do some experiments. Put $n=10,50,100,500,1000,10000,...$  etc and look how the value of the sum approaches. You have to write a program to compute or use online summation-calculator here.

Here, $\displaystyle \mathrm{A}=\lim_{n\to \infty} \frac{1}{n}\left( \frac{n}{n+1}+\frac{n}{n+2}+\frac{n}{n+3}+\cdots+\frac{n}{2n} \right)=\lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n} \left( \frac{n}{n+r} \right)$

It's a limiting sum of which the $n^{\mathrm{th}}$ term of the sequence $\frac{n}{n+1},\frac{n}{n+2},\frac{n}{n+3},\cdots,\frac{n}{n+n}$ converges to exactly $\frac{1}{2}$. So, this sum can be evaluated by a definite integral.

Now $\frac{n}{n+r}=\frac{1}{1+\frac{r}{n}}$. Let $x=\frac{r}{n}$. $\therefore \frac{n}{n+r}=\frac{1}{1+x}\tag{i}$

Notice that for each term i.e. for $r=1,2,3,...$ we have $x=\frac{1}{n},\frac{2}{n},\frac{3}{n},...,\frac{n}{n}$.

As $n \to \infty$, the first term $x=\frac{1}{n} \to 0$. So the initial integral limit is $x=0$. On the other hand, the last term $x=\frac{n}{n}=1$. So the final integral limit is $x=1$.

Now from the continuous set $[0,1]$, for the sake of integration, we need $\Delta x=\frac{1-0}{n}=\frac{1}{n} \tag{ii}$

Now using no$\mathrm{(i)}$ and no$\mathrm{(ii)}$, we obtain

$\displaystyle \mathrm{A}=\lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n} \left( \frac{n}{n+r} \right)=\lim_{n\to \infty} \sum_{r=1}^{n}\Delta x \left( \frac{1}{1+x} \right)=\int_{0}^{1}\frac{1}{1+x}\mathrm{d}x$

Hence \begin{align} \mathrm{A}&=\int_{0}^{1}\frac{1}{1+x}\mathrm{d}x \\ &=[\log_{e}(1+x)]_{0}^{1}\\ &=\log_{e}(2)-\log_{e}(1)\\ &=\log_{e}(2) \end{align}

So the correct answer is C.

by Active (3.5k points)
edited
0
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Can you explain what do you mean by limiting sum?
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How you deduce that the $n^{th}$ term converges to $\frac{1}{2}$
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It's a limiting sum as the series depends on the limiting variable $n$. Here $n\to \infty$, so the series is infinite. I meant $n^{\mathrm{th}}$ term of sequence $\frac{n}{n+1},\frac{n}{n+2},\frac{n}{n+3},\cdots, \frac{n}{n+n}$. In this sequence, the $n^{\mathrm{th}}$ term is $\frac{1}{2}$.
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Ok, thanks!
+2

perfect!

some references :

and other related videos.

$\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty }\sum_{r=1}^{n} f(a+ \Delta x r )\Delta x$

Here, $\lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n} \left( \frac{1}{1+\frac{r}{n}} \right)$

Now, here , $\Delta x = \frac{1}{n}$

and consider $f(x)=\frac{1}{x}$. So, $f(a +\Delta xr) = \frac{1}{a+\frac{r}{n}}$

So, $a$ will be $1$. Now, for $b$,  $\frac{b-a}{n} = \frac{1}{n}$. So, $b$ will be $2$.

So, we have to compute $\int_{1}^{2}\frac{1}{x} dx$ which is  $ln2$

+1
Yeah. 👌
+2
Good resource.
+1 vote
lim n→∞ 1/n(n/n+1+n/n+2+⋯+n/2n)

lim n-> ∞ 1/n( (n+1-1)/(n+1)  + (n+2-2)/(n+2).......(n+n-n)/(2*n))

lim n-> ∞ 1/n( 1 - 1/(n+1) + 1 -2/(n+2) + 1 - 3/(n+3).........1- n/(2*n))     ( here 1 comes n times)

lim n-> ∞ 1/n( n - 1/(n+1)  - 2/(n+2)  - 3/(n+3)......... -n/(2*n))

expand the series, thus

lim n-> ∞ 1 -1/(n*(n+1)) -2/(n*(n+2)).......n/(n*(2*n))

apply lim n-> ∞

we get 1- 1/∞ - 2/∞......n/∞ = 1- 0 - 0.... 0

by (143 points)