Limits

**sinichko** · December 16th 2011, 12:13 PM

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

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

Thanks!

**Siron** · December 16th 2011, 12:42 PM

What have you tried?

1. $n\to ?$

**sinichko** · December 16th 2011, 12:51 PM

Originally Posted by Siron

What have you tried?

1. $n\to ?$

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

1. $n\to \infty$

**Siron** · December 16th 2011, 12:59 PM

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

**chisigma** · December 16th 2011, 06:23 PM

Originally Posted by sinichko

Find limit of
1. $a_{n}=\frac{1}{n-1}+...+\frac{1}{2n}$

Thanks!

Considering that is...

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

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

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

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

Marry Christmas from Serbia

$\chi$ $\sigma$

**sbhatnagar** · December 16th 2011, 07:24 PM

You could also try another approach:

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

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

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

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

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

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

$=\ln(2)$

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

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

**chisigma** · December 16th 2011, 07:50 PM

Originally Posted by sinichko

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

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

Marry Christmas from Serbia

$\chi$ $\sigma$

**Also sprach Zarathustra** · December 17th 2011, 03:42 AM

Originally Posted by Siron

What have you tried?

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

Whrong!

**Siron** · December 17th 2011, 03:53 AM

Originally Posted by Also sprach Zarathustra

Whrong!

You're right.

**besimdervishi** · June 5th 2012, 01:49 AM

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.
please help

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