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n^(1/n)-1-(ln(n))/(n)

Sum of series n^(1/n)-1-(ln(n))/(n)



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The solution

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  oo                      
 ___                      
 \  `                     
  \   /n ___       log(n)\
   )  |\/ n  - 1 - ------|
  /   \              n   /
 /__,                     
n = 1                     
$$\sum_{n=1}^{\infty} \left(\left(n^{\frac{1}{n}} - 1\right) - \frac{\log{\left(n \right)}}{n}\right)$$
Sum(n^(1/n) - 1 - log(n)/n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\left(n^{\frac{1}{n}} - 1\right) - \frac{\log{\left(n \right)}}{n}$$
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
$$a_{n} = n^{\frac{1}{n}} - 1 - \frac{\log{\left(n \right)}}{n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{- n^{\frac{1}{n}} + 1 + \frac{\log{\left(n \right)}}{n}}{- \left(n + 1\right)^{\frac{1}{n + 1}} + 1 + \frac{\log{\left(n + 1 \right)}}{n + 1}}}\right|$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The graph
Sum of series n^(1/n)-1-(ln(n))/(n)

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