# Checking the convergence of two series

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$$>\ln n so a(n) will be decreasing and will tend to 0..but it is just a logic ..Please give me some proper method to check whether it is absolutely convergent or conditionally convergent. ## 2 Answers **hint ** use the fact$$\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}.$$• I get the first one ..but for second one don't get how last inequality will lead to conclusion . – Believer Nov 24 '17 at 20:44 • @omkarGirkar Because$\dfrac{1}{n-\ln n}>\dfrac{1}{n}\$, we don't need that inequality. I am sorry. – Nirvanacs Nov 24 '17 at 20:47
• Ohh..that was easy logic..thanks ... – Believer Nov 24 '17 at 20:51