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Sum of series/ 1/(n^p)

Sum of series 1/(n^p)


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The solution

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

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

    Examples of finding the sum of a series

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