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(-1)^n/nlnn
  • How to use it?

  • Sum of series:
  • (-1)^n/nlnn (-1)^n/nlnn
  • 1/((4k^2)-1) 1/((4k^2)-1)
  • n*0,5^n n*0,5^n
  • (2/9)^n (2/9)^n
  • Identical expressions

  • (- one)^n/nlnn
  • ( minus 1) to the power of n divide by nlnn
  • ( minus one) to the power of n divide by nlnn
  • (-1)n/nlnn
  • -1n/nlnn
  • -1^n/nlnn
  • (-1)^n divide by nlnn
  • Similar expressions

  • (1)^n/nlnn

Sum of series (-1)^n/nlnn



=

The solution

You have entered [src]
  oo              
____              
\   `             
 \        n       
  \   (-1)        
  /   -----*log(n)
 /      n         
/___,             
n = 1             
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n}}{n} \log{\left(n \right)}$$
Sum(((-1)^n/n)*log(n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{n}}{n} \log{\left(n \right)}$$
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
$$a_{n} = \frac{\log{\left(n \right)}}{n}$$
and
$$x_{0} = 1$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(1 + \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left|{\log{\left(n \right)}}\right|}{n \log{\left(n + 1 \right)}}\right)\right)$$
Let's take the limit
we find
False
The rate of convergence of the power series
The answer [src]
  oo              
____              
\   `             
 \        n       
  \   (-1) *log(n)
  /   ------------
 /         n      
/___,             
n = 1             
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n} \log{\left(n \right)}}{n}$$
Sum((-1)^n*log(n)/n, (n, 1, oo))
The graph
Sum of series (-1)^n/nlnn

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