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Sum of series/ (n+2/2n)^n2

Sum of series (n+2/2n)^n2


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The solution

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  oo           
 ___           
 \  `          
  \          n2
  /   (n + n)  
 /__,          
n = 1
$$\sum_{n=1}^{\infty} \left(n + n\right)^{n_{2}}$$ 
Sum((n + n)^n2, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\left(n + n\right)^{n_{2}}$$
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
$$a_{n} = \left(2 n\right)^{n_{2}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(2 n\right)^{\operatorname{re}{\left(n_{2}\right)}} \left(2 n + 2\right)^{- \operatorname{re}{\left(n_{2}\right)}}\right)$$
Let's take the limit
we find
True

False
The answer [src] 
/ n2                       
|2  *zeta(-n2)  for n2 < -1
|                          
|  oo                      
| ___                      
< \  `                     
|  \        n2             
|  /   (2*n)     otherwise 
| /__,                     
|n = 1                     
\
$$\begin{cases} 2^{n_{2}} \zeta\left(- n_{2}\right) & \text{for}\: n_{2} < -1 \\\sum_{n=1}^{\infty} \left(2 n\right)^{n_{2}} & \text{otherwise} \end{cases}$$ 
Piecewise((2^n2*zeta(-n2), n2 < -1), (Sum((2*n)^n2, (n, 1, oo)), True))

    Examples of finding the sum of a series

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