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(1/√n)*ln((n+1)/(n-1))

Sum of series (1/√n)*ln((n+1)/(n-1))



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The solution

You have entered [src]
  oo             
_____            
\    `           
 \        /n + 1\
  \    log|-----|
   \      \n - 1/
   /   ----------
  /        ___   
 /       \/ n    
/____,           
n = 1            
$$\sum_{n=1}^{\infty} \frac{\log{\left(\frac{n + 1}{n - 1} \right)}}{\sqrt{n}}$$
Sum(log((n + 1)/(n - 1))/sqrt(n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\log{\left(\frac{n + 1}{n - 1} \right)}}{\sqrt{n}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \frac{\log{\left(\frac{n + 1}{n - 1} \right)}}{\sqrt{n}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{n + 1} \left|{\log{\left(\frac{n + 1}{n - 1} \right)}}\right|}{\sqrt{n} \log{\left(\frac{n + 2}{n} \right)}}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
Numerical answer [src]
Sum(log((n + 1)/(n - 1))/sqrt(n), (n, 1, oo))
Sum(log((n + 1)/(n - 1))/sqrt(n), (n, 1, oo))
The graph
Sum of series (1/√n)*ln((n+1)/(n-1))

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