0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

**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

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

$\endgroup$
2
$\begingroup$

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 $$\frac{1}{n-\ln n}>\frac{1}{n}.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.