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Sum of series:
n^2/3^n
(n-1)/n!
sqrt(n+1)/(3^(n)*(x+3)^n)
log((n^2+1)/n^2)
Identical expressions
log((n^ two + one)/n^ two)
logarithm of ((n squared plus 1) divide by n squared )
logarithm of ((n to the power of two plus one) divide by n to the power of two)
log((n2+1)/n2)
logn2+1/n2
log((n²+1)/n²)
log((n to the power of 2+1)/n to the power of 2)
logn^2+1/n^2
log((n^2+1) divide by n^2)
Similar expressions
log((n^2-1)/n^2)

Sum of series/ log((n^2+1)/n^2)

Sum of series log((n^2+1)/n^2)

The solution

You have entered [src]

  oo             
____             
\   `            
 \       / 2    \
  \      |n  + 1|
   )  log|------|
  /      |   2  |
 /       \  n   /
/___,            
n = 1

\sum_{n=1}^{\infty} \log{\left(\frac{n^{2} + 1}{n^{2}} \right)}

Sum(log((n^2 + 1)/n^2), (n, 1, oo))

The radius of convergence of the power series

Given number:

\log{\left(\frac{n^{2} + 1}{n^{2}} \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} = \log{\left(\frac{n^{2} + 1}{n^{2}} \right)}

and

x_{0} = 0

d = 0

c = 1

then

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

Let's take the limit
we find

True

False

The rate of convergence of the power series

Numerical answer [src]

1.30184639860371267777043366301

1.30184639860371267777043366301

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 log((n^2+1)/n^2)

The solution

Examples of finding the sum of a series