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(exp^(1/n)-1)^2

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  • Sum of series:
  • 1/(n(n+1)) 1/(n(n+1))
  • (3^(n+3))/(4^(n-1)*5^n) (3^(n+3))/(4^(n-1)*5^n)
  • x^(2*n)
  • n-1 n-1

  • Identical expressions

  • (exp^(one /n)- one)^ two
  • ( exponent of to the power of (1 divide by n) minus 1) squared
  • ( exponent of to the power of (one divide by n) minus one) to the power of two
  • (exp(1/n)-1)2
  • exp1/n-12
  • (exp^(1/n)-1)²
  • (exp to the power of (1/n)-1) to the power of 2
  • exp^1/n-1^2
  • (exp^(1 divide by n)-1)^2

  • Similar expressions

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

Sum of series (exp^(1/n)-1)^2


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The solution

You have entered [src] 
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n = 1
$$\sum_{n=1}^{\infty} \left(e^{\frac{1}{n}} - 1\right)^{2}$$ 
Sum((E^(1/n) - 1)^2, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\left(e^{\frac{1}{n}} - 1\right)^{2}$$
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
$$a_{n} = \left(e^{\frac{1}{n}} - 1\right)^{2}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(e^{\frac{1}{n}} - 1\right)^{2} \left|{\frac{1}{\left(e^{\frac{1}{n + 1}} - 1\right)^{2}}}\right|\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The answer [src] 
  oo            
____            
\   `           
 \             2
  \   /      1\ 
   )  |      -| 
  /   |      n| 
 /    \-1 + e / 
/___,           
n = 1
$$\sum_{n=1}^{\infty} \left(e^{\frac{1}{n}} - 1\right)^{2}$$ 
Sum((-1 + exp(1/n))^2, (n, 1, oo))
Numerical answer [src] 
3.85846821549331100276186447180
3.85846821549331100276186447180
The graph
Sum of series (exp^(1/n)-1)^2

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