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ln(n+1)/(n^(5/4))
  • How to use it?

  • Sum of series:
  • 9/(9n^2+21n-8) 9/(9n^2+21n-8)
  • (3-sin*n)/n-lnn (3-sin*n)/n-lnn
  • 1/(1+n^2) 1/(1+n^2)
  • e^(-n) e^(-n)
  • Identical expressions

  • ln(n+ one)/(n^(five / four))
  • ln(n plus 1) divide by (n to the power of (5 divide by 4))
  • ln(n plus one) divide by (n to the power of (five divide by four))
  • ln(n+1)/(n(5/4))
  • lnn+1/n5/4
  • lnn+1/n^5/4
  • ln(n+1) divide by (n^(5 divide by 4))
  • Similar expressions

  • ln(n-1)/(n^(5/4))

Sum of series ln(n+1)/(n^(5/4))



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The solution

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  oo            
____            
\   `           
 \    log(n + 1)
  \   ----------
  /       5/4   
 /       n      
/___,           
n = 1           
$$\sum_{n=1}^{\infty} \frac{\log{\left(n + 1 \right)}}{n^{\frac{5}{4}}}$$
Sum(log(n + 1)/n^(5/4), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\log{\left(n + 1 \right)}}{n^{\frac{5}{4}}}$$
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
$$a_{n} = \frac{\log{\left(n + 1 \right)}}{n^{\frac{5}{4}}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{\frac{5}{4}} \log{\left(n + 1 \right)}}{n^{\frac{5}{4}} \log{\left(n + 2 \right)}}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The graph
Sum of series ln(n+1)/(n^(5/4))

    Examples of finding the sum of a series