Mister Exam

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How to use it?
Sum of series:
sqrt(n+1)-sqrt(n)/sqrt(n^2+n)
(3*x^2-2*x+3)/factorial(x+1)
1/ln(5^n)
1/
Identical expressions
(twelve *ln(x))/(x^ three)
(12 multiply by ln(x)) divide by (x cubed )
(twelve multiply by ln(x)) divide by (x to the power of three)
(12*ln(x))/(x3)
12*lnx/x3
(12*ln(x))/(x³)
(12*ln(x))/(x to the power of 3)
(12ln(x))/(x^3)
(12ln(x))/(x3)
12lnx/x3
12lnx/x^3
(12*ln(x)) divide by (x^3)

Sum of series/ (12*ln(x))/(x^3)

Sum of series (12*ln(x))/(x^3)

The solution

You have entered [src]

  oo           
____           
\   `          
 \    12*log(x)
  \   ---------
  /        3   
 /        x    
/___,          
n = 1

\sum_{n=1}^{\infty} \frac{12 \log{\left(x \right)}}{x^{3}}

Sum((12*log(x))/x^3, (n, 1, oo))

The radius of convergence of the power series

Given number:

\frac{12 \log{\left(x \right)}}{x^{3}}

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{12 \log{\left(x \right)}}{x^{3}}

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)
---------
     3   
    x

\frac{\infty \log{\left(x \right)}}{x^{3}}

oo*log(x)/x^3

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
```