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

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  • Sum of series:
  • (n^3+n+5)/(n+2) (n^3+n+5)/(n+2)
  • (0.15/0.05)^1.3 (0.15/0.05)^1.3
  • 5/(25*n^2-5*n-6) 5/(25*n^2-5*n-6)
  • tan(sqrt(n)/(n^2+1)) tan(sqrt(n)/(n^2+1))

  • Identical expressions

  • tan(sqrt(n)/(n^ two + one))
  • tangent of ( square root of (n) divide by (n squared plus 1))
  • tangent of ( square root of (n) divide by (n to the power of two plus one))
  • tan(√(n)/(n^2+1))
  • tan(sqrt(n)/(n2+1))
  • tansqrtn/n2+1
  • tan(sqrt(n)/(n²+1))
  • tan(sqrt(n)/(n to the power of 2+1))
  • tansqrtn/n^2+1
  • tan(sqrt(n) divide by (n^2+1))

  • Similar expressions

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

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


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The solution

You have entered [src] 
  oo             
____             
\   `            
 \       /  ___ \
  \      |\/ n  |
   )  tan|------|
  /      | 2    |
 /       \n  + 1/
/___,            
n = 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:
$$\tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}$$
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
$$a_{n} = \tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\tan{\left(\frac{\sqrt{n}}{n^{2} + 1} \right)}}{\tan{\left(\frac{\sqrt{n + 1}}{\left(n + 1\right)^{2} + 1} \right)}}}\right|$$
Let's take the limit
we find
True

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

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