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(arctg(1/sqrt(n-1)))/(n-1)
Sum of series/ (arctg(1/sqrt(n-1)))/(n-1)

Sum of series (arctg(1/sqrt(n-1)))/(n-1)


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The solution

You have entered [src] 
  oo                  
_____                 
\    `                
 \         /    1    \
  \    atan|---------|
   \       |  _______|
   /       \\/ n - 1 /
  /    ---------------
 /          n - 1     
/____,                
n = 1
∑n=1∞atan⁡(1n−1)n−1\sum_{n=1}^{\infty} \frac{\operatorname{atan}{\left(\frac{1}{\sqrt{n - 1}} \right)}}{n - 1} 
Sum(atan(1/(sqrt(n - 1)))/(n - 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
atan⁡(1n−1)n−1\frac{\operatorname{atan}{\left(\frac{1}{\sqrt{n - 1}} \right)}}{n - 1}
It is a series of species
an(cx−x0)dna_{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:
Rd=x0+lim⁡n→∞∣anan+1∣cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=atan⁡(1n−1)n−1a_{n} = \frac{\operatorname{atan}{\left(\frac{1}{\sqrt{n - 1}} \right)}}{n - 1}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞(n∣atan⁡(1n−1)n−1∣atan⁡(1n))1 = \lim_{n \to \infty}\left(\frac{n \left|{\frac{\operatorname{atan}{\left(\frac{1}{\sqrt{n - 1}} \right)}}{n - 1}}\right|}{\operatorname{atan}{\left(\frac{1}{\sqrt{n}} \right)}}\right)
Let's take the limit
we find
True

False
The rate of convergence of the power series
-0.010-0.008-0.006-0.004-0.0020.0100.0000.0020.0040.0060.0080.00
The answer [src] 
  oo                   
_____                  
\    `                 
 \         /    1     \
  \    atan|----------|
   \       |  ________|
   /       \\/ -1 + n /
  /    ----------------
 /          -1 + n     
/____,                 
n = 1
∑n=1∞atan⁡(1n−1)n−1\sum_{n=1}^{\infty} \frac{\operatorname{atan}{\left(\frac{1}{\sqrt{n - 1}} \right)}}{n - 1} 
Sum(atan(1/sqrt(-1 + n))/(-1 + n), (n, 1, oo))
Numerical answer [src] 
Sum(atan(1/(sqrt(n - 1)))/(n - 1), (n, 1, oo))
Sum(atan(1/(sqrt(n - 1)))/(n - 1), (n, 1, oo))
The graph
Sum of series (arctg(1/sqrt(n-1)))/(n-1)

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