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

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


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The solution

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  oo            
____            
\   `           
 \        n + 1 
  \   (-1)      
  /   ----------
 /    log(n + 2)
/___,           
n = 1
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{\log{\left(n + 2 \right)}}$$ 
Sum((-1)^(n + 1)/log(n + 2), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{n + 1}}{\log{\left(n + 2 \right)}}$$
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
$$a_{n} = \frac{\left(-1\right)^{n + 1}}{\log{\left(n + 2 \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\log{\left(n + 3 \right)}}{\log{\left(n + 2 \right)}}\right)$$
Let's take the limit
we find
True

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

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