Mister Exam

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Sum of series:
sqrt(n)-2*sqrt(n+1)+sqrt(n+2)
sqrt(1+1)/1
x^n*4^n/((5^n*(n+9)^(1/9)))
197^(197*230)*10^(20*(197\230))
Identical expressions
two ^2x(x- one / four)x
2 squared x(x minus 1 divide by 4)x
two squared x(x minus one divide by four)x
22x(x-1/4)x
22xx-1/4x
2²x(x-1/4)x
2 to the power of 2x(x-1/4)x
2^2xx-1/4x
2^2x(x-1 divide by 4)x
Similar expressions
2^2x(x+1/4)x

Sum of series/ 2^2x(x-1/4)x

Sum of series 2^2x(x-1/4)x

The solution

You have entered [src]

  oo                 
 __                  
 \ `                 
  )   4*x*(x - 1/4)*x
 /_,                 
n = 1

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

Sum(((4*x)*(x - 1/4))*x, (n, 1, oo))

The radius of convergence of the power series

Given number:

x 4 x \left(x - \frac{1}{4}\right)

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} = 4 x^{2} \left(x - \frac{1}{4}\right)

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]

    2           
oo*x *(-1/4 + x)

\infty x^{2} \left(x - \frac{1}{4}\right)

oo*x^2*(-1/4 + x)

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
```