# Thread: Proving Limit

1. ## Proving Limit

Prove that $lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]=1$

I am thinking something like that :
$lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]= lim[\frac{n^2}{n^2(n+\frac{1}{n^2})}+\frac{n^2}{n^2(n+ \frac{2}{n^2})}+...+\frac{n^2}{n^2(n+\frac{1}{n})}]$ and since $\frac{1}{n^2}\rightarrow0$ we get
$lim[\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}]=1$

Am i right?

2. What value are you making $\displaystyle n$ approach?

3. Originally Posted by Prove It
What value are you making $\displaystyle n$ approach?
$n\rightarrow\infty$

4. Originally Posted by SENTINEL4
Prove that $lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]=1$

I am thinking something like that :
$lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]= lim[\frac{n^2}{n^2(n+\frac{1}{n^2})}+\frac{n^2}{n^2(n+ \frac{2}{n^2})}+...+\frac{n^2}{n^2(n+\frac{1}{n})}]$ and since $\frac{1}{n^2}\rightarrow0$ we get
$lim[\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}]=1$

Am i right?
According to me your method is wrong...
instead take (n+1/(n^2))...... in the numerator with an exponent of '-1' and take 'n' common from each bracket and then expand the bracket binomially....
now apply the limit n->infinity..you should get the answer..

5. Just because a term "goes to 0", that does't mean you can ignore it in the sum. Its values for finite n are NOT 0 and will affect the sum. For example, 1/n goes to 0 but $\sum\frac{1}{n}$ goes to infinity.

6. Which is the correct method to reach to the answer then?

7. I don't think this limit does go to $\displaystyle 1$...

The limit of a sum is the sum of the limits, and if you evaluate the limit of each term using L'Hospital's Rule, you should find each term $\displaystyle \to 0$.

So the limit should be $\displaystyle 0$...

8. Originally Posted by SENTINEL4
Prove that $lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]=1$
That's exactly what the exercise in the book says.

9. The problem is to find [if it exists...] the limit of the sequence...

$\displaystyle s_{n}= \sum_{k=1}^{n} a_{k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n^{2}}} = \sum_{k=1}^{n} \frac{1}{n} (1-\frac{k}{n^{3}+k})$ (1)

It is evident from (1) that, because $s_{n}$ is the sum of n positive terms and for all k is $a_{k}< \frac{1}{n}$ for all n will be $s_{n}<1$. But for all k is also...

$\displaystyle \lim_{n \rightarrow \infty} n\ a_{k}=1$ (2)

... so that...

$\displaystyle \lim_{n \rightarrow \infty} s_{n}=1$ (3)

The first ten $s_{n}$ are...

$s_{1}= .5$

$s_{2}= .844444444444...$

$s_{3}= .931773399015...$

$s_{4}= .962678176374...$

$s_{5}= .976681734034...$

$s_{6}= .984114247235...$

$s_{7}= .988505461555...$

$s_{8}= .991307021026...$

$s_{9}= .993200301061...$

$s_{10}= .994538200011...$

Kind regards

$\chi$ $\sigma$

10. If $L$ is the limit of $(a_n)$ (the given sequence) we have:

$\dfrac{n^2}{n^3+n}+\ldots+\dfrac{n^2}{n^3+n}\leq a_n\leq \dfrac{n^2}{n^3}+\ldots+\dfrac{n^2}{n^3}$

Equivalently:

$\dfrac{n^3}{n^3+n}\leq a_n\leq \dfrac{n^3}{n^3}$

Taking limits:

$1\leq L\leq 1$

which implies $L=1$ .

Fernando Revilla

11. Originally Posted by Prove It
The limit of a sum is the sum of the limits ...

Here the result is not valid because the number of terms of $a_n$ depends on $n$ .

Fernando Revilla

12. Originally Posted by Prove It
I don't think this limit does go to $\displaystyle 1$...

The limit of a sum is the sum of the limits, and if you evaluate the limit of each term using L'Hospital's Rule, you should find each term $\displaystyle \to 0$.

So the limit should be $\displaystyle 0$...
Here we cannot apply L'Hopital's rule since the function is not continuous as it takes only integer values .the function will have discreet set of points.
For ex:if 1<=n<=2 it will take only n=1 and n=2 but not the real nos. b/w 1&2