Mister Exam

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Sum of series:
(4*9^n-2^n)/18^n
1/((3*n+1)*(3*n+4))
sqrt(n)*(x+3)^n/n^2+1
ln((n(n+2))/(n+1)^2)
Identical expressions
ln(x)/(x^(five / four))
ln(x) divide by (x to the power of (5 divide by 4))
ln(x) divide by (x to the power of (five divide by four))
ln(x)/(x(5/4))
lnx/x5/4
lnx/x^5/4
ln(x) divide by (x^(5 divide by 4))

Sum of series ln(x)/(x^(5/4))

The solution

You have entered [src]

  oo        
____        
\   `       
 \    log(x)
  \   ------
  /     5/4 
 /     x    
/___,       
n = 1

$$\sum_{n=1}^{\infty} \frac{\log{\left(x \right)}}{x^{\frac{5}{4}}}$$

Sum(log(x)/x^(5/4), (n, 1, oo))

The radius of convergence of the power series

Given number:
$$\frac{\log{\left(x \right)}}{x^{\frac{5}{4}}}$$
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
$$a_{n} = \frac{\log{\left(x \right)}}{x^{\frac{5}{4}}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find

True

False

The answer [src]

oo*log(x)
---------
    5/4  
   x

$$\frac{\infty \log{\left(x \right)}}{x^{\frac{5}{4}}}$$

oo*log(x)/x^(5/4)

Examples of finding the sum of a series

The Sum of the Power Series
```
x^n/n
```
```
(x-1)^n
```
Factorial
```
1/2^(n!)
```
```
n^2/n!
```
```
x^n/n!
```
```
k!/(n!*(n+k)!)
```
Flint Hills Series
```
csc(n)^2/n^3
```
Basel problem
```
1/n^2
```
```
1/n^4
```
```
1/n^6
```
Harmonic series
```
1/n
```
Grandi's series
```
(-1)^n
```
Alternating series
```
(-1)^(n + 1)/n
```
```
(n + 2)*(-1)^(n - 1)
```
```
(3*n - 1)/(-5)^n
```
```
(-1)^(n - 1)*n/(6*n - 5)
```
Newton–Mercator series
```
(-1)^(n + 1)/n*x^n
```
Exam the series for convergence
```
(3*n - 1)/(-5)^n
```

Other calculators

How to use it?

Identical expressions

Sum of series ln(x)/(x^(5/4))

The solution

Examples of finding the sum of a series