I am totally blank about this two series.

1.$\sum_{n=2}^{\infty}\left(\frac{1}{n^{1/2}}\right)\ln\left(\frac{n+1}{n-1}\right)$...For this series I just have one logic that it is product of two divergent series i.e.first is $\sum (1/n^{1/2})$ and the remaining one..so I think the original series should diverge..but I doubt on this..

2.$\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n-\ln n}\right)$..I try to find whether it is absolutely convergent but with no conclusion.I feel that this series might be convergent by alternating series test because $n$$>\ln n$ so $a(n)$ will be decreasing and will tend to $0$..but it is just a logic ..Please give me some proper method to check whether it is absolutely convergent or conditionally convergent.


**hint **

use the fact

$$\frac {n+1}{n-1}=1+\frac {2}{n-1} $$

and near $\infty $, $$\ln (\frac {n+1}{n-1})\sim \frac{2}{n} $$

thus $$u_n\sim \frac {2}{n^\frac 32} $$

and $\sum u_n $ converges.

for the second observe that

$$n-\ln (n)\sim n $$


Note that $\ln (1+x)\sim x$ for $x$ sufficiently small. Then $$\ln\Big(\frac{n+1}{n-1}\Big)=\ln\Big(1+\frac{2}{n-1}\Big)\sim \frac{2}{n-1}$$Hence it absolutely convergent.

The second one is just conditionally convergent. Since you can show that $$\frac{1}{n-\ln n}>\frac{1}{n}.$$

  • $\begingroup$ I get the first one ..but for second one don't get how last inequality will lead to conclusion . $\endgroup$ – Believer Nov 24 '17 at 20:44
  • $\begingroup$ @omkarGirkar Because $\dfrac{1}{n-\ln n}>\dfrac{1}{n}$, we don't need that inequality. I am sorry. $\endgroup$ – Nirvanacs Nov 24 '17 at 20:47
  • $\begingroup$ Ohh..that was easy logic..thanks ... $\endgroup$ – Believer Nov 24 '17 at 20:51

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