$\displaystyle 1-\frac{1}{3}\left(1+\frac{1}{2}\right)+\frac{1}{5}\ left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{7}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1 }{4}\right)+...$

show that the above series is converges(Rofl)

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

show that the above series is converges(Rofl) - Nov 26th 2009, 08:03 PMRockHard
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 PMXingyuan
You probably mean this : $\displaystyle (-1)^{2n-1}\frac{\ln n}{2n-1}$

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

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

... where...

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

It is well known the limit...

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

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

$\displaystyle \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 $\displaystyle n \rightarrow \infty$. Therefore (1) is convergent...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 26th 2009, 08:52 PMRockHard
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 PMmathaddict
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

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

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 .

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

Putting them together :

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

Notice that as n approaches infinity , $\displaystyle \frac{1}{2n-1}(-1)^{n+1}$ would approach 0 so that this series would eventually converge to 0 . - Nov 26th 2009, 10:31 PMRockHard
- Nov 27th 2009, 08:21 AMmathaddict