Mister Exam

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Sum of series:
ln((n(n+2))/(n+1)^2)
2(x-6)^(4n)/(n(16)^n)
sin(3n)/(7n)^(1/5)
(-1)^(n-1)/ln(n+1)
Graphing y =:
k^2-5*k+6
Identical expressions
k^ two - five *k+ six
k squared minus 5 multiply by k plus 6
k to the power of two minus five multiply by k plus six
k2-5*k+6
k²-5*k+6
k to the power of 2-5*k+6
k^2-5k+6
k2-5k+6
Similar expressions
k^2+5*k+6
k^2-5*k-6

Sum of series/ k^2-5*k+6

Sum of series k^2-5*k+6

The solution

You have entered [src]

  oo                
 ___                
 \  `               
  \   / 2          \
  /   \k  - 5*k + 6/
 /__,               
n = 1

\sum_{n=1}^{\infty} \left(\left(k^{2} - 5 k\right) + 6\right)

Sum(k^2 - 5*k + 6, (n, 1, oo))

The radius of convergence of the power series

Given number:

\left(k^{2} - 5 k\right) + 6

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} = k^{2} - 5 k + 6

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*\6 + k  - 5*k/

\infty \left(k^{2} - 5 k + 6\right)

oo*(6 + k^2 - 5*k)

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
```