1. ## Limits

Help me please with 2 questions:
Find limit of
1.$\displaystyle a_{n}=\frac{1}{n-1}+...+\frac{1}{2n}$

2.$\displaystyle \underset{n \to \infty }{\lim} \left [ \frac{1}{n^{2}}+...+\frac{1}{(2n)^{2}} \right ]$

Thanks!

2. ## Re: Limits

What have you tried?

1. $\displaystyle n\to ?$

3. ## Re: Limits

Originally Posted by Siron
What have you tried?

1. $\displaystyle n\to ?$

2. If $\displaystyle n\to \infty$ then $\displaystyle \frac{1}{n^2}\to 0$, same for the other terms.

1. $\displaystyle n\to \infty$

4. ## Re: Limits

The limit of a sum is the sum of the limits of the separated terms.

5. ## Re: Limits

Originally Posted by sinichko
Find limit of
1.$\displaystyle a_{n}=\frac{1}{n-1}+...+\frac{1}{2n}$

Thanks!
Considering that is...

$\displaystyle \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n-2} \frac{1}{k} - \ln (n-2) \} =\gamma$ (1)

$\displaystyle \lim_{n \rightarrow \infty} \{\sum_{k=1}^{2n} \frac{1}{k} - \ln 2 n \} =\gamma$ (2)

... subtracting (1) from (2) You obtain...

$\displaystyle \lim_{n \rightarrow \infty} \sum_{k=n-1}^{2n} \frac{1}{k} = \lim_{n \rightarrow \infty} \{\ln 2 n - \ln (n-2)\}= \ln 2$ (3)

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$

6. ## Re: Limits

You could also try another approach:

$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}\right]$
$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}+\cdots +\frac{1}{2n}\right]$

$\displaystyle = \left[ \lim_{n \to \infty}\frac{1}{n-1}+\lim_{n \to \infty}\frac{1}{n}+ \lim_{n \to \infty}\sum_{r=1}^{n}\frac{1}{n+r}\right]$

$\displaystyle = \left[ 0+0+\lim_{n \to \infty}\sum_{r=1}^{n}\left(\frac{1}{n} \right)\frac{1}{1+{r \over n}}\right]$

$\displaystyle =\int_{0}^{1} \frac{1}{1+x} dx$

$\displaystyle =[\ln(1+x)]^1_0$

$\displaystyle =\ln(2)$

$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n^2}+\cdots +\frac{1}{(2n)^2}\right]$
$\displaystyle \lim_{n \to \infty} \left[ \frac{1}{n^2}+\cdots +\frac{1}{(2n)^2}\right]=0$

7. ## Re: Limits

Originally Posted by sinichko
2.$\displaystyle \underset{n \to \infty }{\lim} \left [ \frac{1}{n^{2}}+...+\frac{1}{(2n)^{2}} \right ]$

Defining $\displaystyle a_{n}= \frac{1}{n^{2}}+...+\frac{1}{(2 n)^{2}}$ it is evident that $\displaystyle 0<a_{n}<\frac{1}{n}$ so that is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$...

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$

8. ## Re: Limits

Originally Posted by Siron
What have you tried?

2. If $\displaystyle n\to \infty$ then $\displaystyle \frac{1}{n^2}\to 0$, same for the other terms.
Whrong!

9. ## Re: Limits

Originally Posted by Also sprach Zarathustra
Whrong!
You're right.

10. ## Re: Limits

need some help in this please!!
is given the function {y=sin2x/x for x different from 0} or {y=2a for x=o}. find "a" if this function is continuous everywhere.