$\displaystyle 1-\frac{1}{3}\left(1+\frac{1}{2}\right)+\frac{1}{5}\ left(1+\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{7}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1 }{4}\right)+...$
show that the above series is converges
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
$\displaystyle \chi$ $\displaystyle \sigma$
Aw, man I was trying to figure out the function for it, glad I started 5 minutes ago
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
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 .