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  • Sum of series:
  • 1/((1+r)^n)
  • (1)/(2^n+n) (1)/(2^n+n)
  • e^2 e^2
  • -13 -13

  • Identical expressions

  • one /((one +r)^n)
  • 1 divide by ((1 plus r) to the power of n)
  • one divide by ((one plus r) to the power of n)
  • 1/((1+r)n)
  • 1/1+rn
  • 1/1+r^n
  • 1 divide by ((1+r)^n)

  • Similar expressions

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

Sum of series 1/((1+r)^n)


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The solution

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

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