Mister Exam

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Sum of series:
6/9n^2+12n-5
n^n/2^(n+1)
e^(-n)
1/((3n+1)*(3n+4))
Identical expressions
(one /√n)*ln((n+ one)/(n- one))
(1 divide by √n) multiply by ln((n plus 1) divide by (n minus 1))
(one divide by √n) multiply by ln((n plus one) divide by (n minus one))
(1/√n)ln((n+1)/(n-1))
1/√nlnn+1/n-1
(1 divide by √n)*ln((n+1) divide by (n-1))
Similar expressions
(1/√n)*ln((n+1)/(n+1))
(1/√n)*ln((n-1)/(n-1))

Sum of series/ (1/√n)*ln((n+1)/(n-1))

Sum of series (1/√n)*ln((n+1)/(n-1))

The solution

You have entered [src]

  oo             
_____            
\    `           
 \        /n + 1\
  \    log|-----|
   \      \n - 1/
   /   ----------
  /        ___   
 /       \/ n    
/____,           
n = 1

\sum_{n=1}^{\infty} \frac{\log{\left(\frac{n + 1}{n - 1} \right)}}{\sqrt{n}}

Sum(log((n + 1)/(n - 1))/sqrt(n), (n, 1, oo))

The radius of convergence of the power series

Given number:

\frac{\log{\left(\frac{n + 1}{n - 1} \right)}}{\sqrt{n}}

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(\frac{n + 1}{n - 1} \right)}}{\sqrt{n}}

and

x_{0} = 0

d = 0

c = 1

then

1 = \lim_{n \to \infty}\left(\frac{\sqrt{n + 1} \left|{\log{\left(\frac{n + 1}{n - 1} \right)}}\right|}{\sqrt{n} \log{\left(\frac{n + 2}{n} \right)}}\right)

Let's take the limit
we find

True

False

The rate of convergence of the power series

Numerical answer [src]

Sum(log((n + 1)/(n - 1))/sqrt(n), (n, 1, oo))

Sum(log((n + 1)/(n - 1))/sqrt(n), (n, 1, oo))

The graph

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
```

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How to use it?

Identical expressions

Similar expressions

Sum of series (1/√n)*ln((n+1)/(n-1))

The solution

Examples of finding the sum of a series