Series converge

• Nov 26th 2009, 07:53 PM
Xingyuan
Series converge
$1-\frac{1}{3}\left(1+\frac{1}{2}\right)+\frac{1}{5}\ left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{7}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1 }{4}\right)+...$

show that the above series is converges(Rofl)
• Nov 26th 2009, 08:03 PM
RockHard
Looks like somewhat are alternating series to me, find a function to which to fit the pattern of the terms you were given and simply use the alternating series test if you can
• Nov 26th 2009, 08:35 PM
Xingyuan
You probably mean this : $(-1)^{2n-1}\frac{\ln n}{2n-1}$

(Giggle)
• Nov 26th 2009, 08:40 PM
chisigma
The series can be written in compact form as...

$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{n}}{2n-1}$ (1)

... where...

$H_{n} = \sum_{k=1}^{n} \frac{1}{k}$ (2)

It is well known the limit...

$\lim_{n \rightarrow \infty} H_{n} - \ln n = \gamma$ (3)

... where $\gamma$ is the 'Euler's constant'. Consequence of (3) is...

$\lim_{n \rightarrow \infty} \frac{H_{n}}{2n-1} = \lim_{n \rightarrow \infty} \frac{\ln n + \gamma}{2n-1} = 0$ (4)

... so that the (1) is an alternating decreasing terms series with the general term that in absolute value tends to 0 if $n \rightarrow \infty$. Therefore (1) is convergent...

Kind regards

$\chi$ $\sigma$
• Nov 26th 2009, 08:52 PM
RockHard
Aw, man I was trying to figure out the function for it, glad I started 5 minutes ago (Wink)
That is a new version of the alternating series I have not worked on yet so thank you for the question and thank you to chisigma for the answer.
Holy, that is a very new series in this form for me I would have never got natural log, how do you know when to apply such things chisigma
• Nov 26th 2009, 10:26 PM
Quote:

Originally Posted by Xingyuan
$1-\frac{1}{3}\left(1+\frac{1}{2}\right)+\frac{1}{5}\ left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{7}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1 }{4}\right)+...$

show that the above series is converges(Rofl)

HI

You can break these to find the general term .

1/1-1/3+1/5-1/7+... notice that the denominators are odd and the signs alternate so it would be

$\sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}]$

The other part :

(1)+(1+1/2)+(1+1/2+1/3)+... The pattern should be obvious here . Every subsequent fraction added has its denominator equal to the no of terms .

$\sum^{\infty}_{n=1}[\sum^{n}_{r=1}(\frac{1}{r})]$

Putting them together :

$\sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}\cdot \sum^{n}_{r=1}(\frac{1}{r})]$

Notice that as n approaches infinity , $\frac{1}{2n-1}(-1)^{n+1}$ would approach 0 so that this series would eventually converge to 0 .
• Nov 26th 2009, 10:31 PM
RockHard
Quote:

HI

You can break these to find the general term .

1/1-1/3+1/5-1/7+... notice that the denominators are odd and the signs alternate so it would be

$\sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}]$

The other part :

(1)+(1+1/2)+(1+1/2+1/3)+... The pattern should be obvious here . Every subsequent fraction added has its denominator equal to the no of terms .

$\sum^{\infty}_{r=1}[\sum^{n}_{r=1}(\frac{1}{n})]$

Putting them together :

$\sum^{\infty}_{n=1}[\frac{1}{2n-1}(-1)^{n+1}\cdot \sum^{n}_{r=1}(\frac{1}{r})]$

Notice that as n approaches infinity , $\frac{1}{2n-1}(-1)^{n+1}$ would approach 0 so that this series would eventually converge to 0 .

Why right here would you say the to the nth term not $\infty$

$\sum^{\infty}_{r=1}[\sum^{n}_{r=1}(\frac{1}{n})]$

Just curious I am still fairly new to this
• Nov 27th 2009, 08:21 AM
Quote:

Originally Posted by RockHard
Why right here would you say the to the nth term not $\infty$

$\sum^{\infty}_{r=1}[\sum^{n}_{r=1}(\frac{1}{n})]$

Just curious I am still fairly new to this

HI

My bad . It should be

$\sum^{\infty}_{n=1}[\sum^{n}_{r=1}(\frac{1}{r})]$

$\sum^{\infty}_{n=1}=\sum^{1}_{r=1}(\frac{1}{r})+\s um^{2}(\frac{1}{r})+\sum^{3}_{r=1}(\frac{1}{r})+.. .$