# Proving Limit

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• Jan 25th 2011, 02:30 AM
SENTINEL4
Proving Limit
Prove that $\displaystyle lim[\frac{n^2}{n^3+1}+\frac{n^2}{n^3+2}+...+\frac{n^2} {n^3+n}]=1$

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

Am i right?
• Jan 25th 2011, 02:43 AM
Prove It
What value are you making $\displaystyle \displaystyle n$ approach?
• Jan 25th 2011, 02:49 AM
SENTINEL4
Quote:

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

$\displaystyle n\rightarrow\infty$
• Jan 25th 2011, 03:38 AM
themekanikal
Quote:

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

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

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..
• Jan 25th 2011, 05:00 AM
HallsofIvy
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 $\displaystyle \sum\frac{1}{n}$ goes to infinity.
• Jan 25th 2011, 05:49 AM
SENTINEL4
Which is the correct method to reach to the answer then?
• Jan 25th 2011, 06:04 AM
Prove It
I don't think this limit does go to $\displaystyle \displaystyle 1$...

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 $\displaystyle \displaystyle \to 0$.

So the limit should be $\displaystyle \displaystyle 0$...
• Jan 25th 2011, 06:08 AM
SENTINEL4
Quote:

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

That's exactly what the exercise in the book says.
• Jan 25th 2011, 08:09 AM
chisigma
The problem is to find [if it exists...] the limit of the sequence...

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

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

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

... so that...

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

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

$\displaystyle s_{1}= .5$

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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jan 25th 2011, 09:18 AM
FernandoRevilla
If $\displaystyle L$ is the limit of $\displaystyle (a_n)$ (the given sequence) we have:

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

Equivalently:

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

Taking limits:

$\displaystyle 1\leq L\leq 1$

which implies $\displaystyle L=1$ .

Fernando Revilla
• Jan 25th 2011, 10:06 AM
FernandoRevilla
Quote:

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 $\displaystyle a_n$ depends on $\displaystyle n$ .

Fernando Revilla
• Jan 25th 2011, 09:50 PM
themekanikal
Quote:

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

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 $\displaystyle \displaystyle \to 0$.

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

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