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tan(sqrt(n)/(n^2+1))

Sum of series tan(sqrt(n)/(n^2+1))



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The solution

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  oo             
____             
\   `            
 \       /  ___ \
  \      |\/ n  |
   )  tan|------|
  /      | 2    |
 /       \n  + 1/
/___,            
n = 1            
n=1tan(nn2+1)\sum_{n=1}^{\infty} \tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}
Sum(tan(sqrt(n)/(n^2 + 1)), (n, 1, oo))
The radius of convergence of the power series
Given number:
tan(nn2+1)\tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}
It is a series of species
an(cxx0)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+limnanan+1cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=tan(nn2+1)a_{n} = \tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=limntan(nn2+1)tan(n+1(n+1)2+1)1 = \lim_{n \to \infty} \left|{\frac{\tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}}{\tan{\left(\frac{\sqrt{n + 1}}{\left(n + 1\right)^{2} + 1} \right)}}}\right|
Let's take the limit
we find
True

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.02.0
The graph
Sum of series tan(sqrt(n)/(n^2+1))

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