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  • Sum of series:
  • 2^n 2^n
  • ln(n+1)/(n+1) ln(n+1)/(n+1)
  • (n*x^n)/(n+1)
  • yi
  • Identical expressions

  • (n*x^n)/(n+ one)
  • (n multiply by x to the power of n) divide by (n plus 1)
  • (n multiply by x to the power of n) divide by (n plus one)
  • (n*xn)/(n+1)
  • n*xn/n+1
  • (nx^n)/(n+1)
  • (nxn)/(n+1)
  • nxn/n+1
  • nx^n/n+1
  • (n*x^n) divide by (n+1)
  • Similar expressions

  • (n*x^n)/(n-1)

Sum of series (n*x^n)/(n+1)



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The solution

You have entered [src]
  oo       
____       
\   `      
 \        n
  \    n*x 
  /   -----
 /    n + 1
/___,      
n = 0      
$$\sum_{n=0}^{\infty} \frac{n x^{n}}{n + 1}$$
Sum((n*x^n)/(n + 1), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\frac{n x^{n}}{n + 1}$$
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
$$a_{n} = \frac{n}{n + 1}$$
and
$$x_{0} = 0$$
,
$$d = 1$$
,
$$c = 1$$
then
$$R = \lim_{n \to \infty}\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}}\right)$$
Let's take the limit
we find
$$R = 1$$
The answer [src]
/  /    2      2*log(1 - x)\             
|x*|- ------ + ------------|             
|  |   2             2     |             
|  \  x  - x        x      /             
|---------------------------  for |x| < 1
|             2                          
|                                        
|          oo                            
<        ____                            
|        \   `                           
|         \        n                     
|          \    n*x                      
|          /   -----           otherwise 
|         /    1 + n                     
|        /___,                           
|        n = 0                           
\                                        
$$\begin{cases} \frac{x \left(- \frac{2}{x^{2} - x} + \frac{2 \log{\left(1 - x \right)}}{x^{2}}\right)}{2} & \text{for}\: \left|{x}\right| < 1 \\\sum_{n=0}^{\infty} \frac{n x^{n}}{n + 1} & \text{otherwise} \end{cases}$$
Piecewise((x*(-2/(x^2 - x) + 2*log(1 - x)/x^2)/2, |x| < 1), (Sum(n*x^n/(1 + n), (n, 0, oo)), True))

    Examples of finding the sum of a series