Mister Exam

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Sum of series:
(3n+4)/(n/((n+1)(n+2)))
e^n/(1+e^(2*n))
1/((2n-1)(2n-1))
(3-2i-5i^2+8i^3)
Identical expressions
e^n/(one +e^(two *n))
e to the power of n divide by (1 plus e to the power of (2 multiply by n))
e to the power of n divide by (one plus e to the power of (two multiply by n))
en/(1+e(2*n))
en/1+e2*n
e^n/(1+e^(2n))
en/(1+e(2n))
en/1+e2n
e^n/1+e^2n
e^n divide by (1+e^(2*n))
Similar expressions
e^n/(1-e^(2*n))

Sum of series/ e^n/(1+e^(2*n))

Sum of series e^n/(1+e^(2*n))

The solution

You have entered [src]

  oo          
____          
\   `         
 \        n   
  \      E    
   )  --------
  /        2*n
 /    1 + E   
/___,         
n = 1

$$\sum_{n=1}^{\infty} \frac{e^{n}}{e^{2 n} + 1}$$

Sum(E^n/(1 + E^(2*n)), (n, 1, oo))

The radius of convergence of the power series

Given number:
$$\frac{e^{n}}{e^{2 n} + 1}$$
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
$$a_{n} = \frac{1}{e^{2 n} + 1}$$
and
$$x_{0} = - e$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(- e + \lim_{n \to \infty}\left(\frac{e^{2 n + 2} + 1}{e^{2 n} + 1}\right)\right)$$
Let's take the limit
we find

False

False

The rate of convergence of the power series

The answer [src]

  oo          
____          
\   `         
 \        n   
  \      e    
   )  --------
  /        2*n
 /    1 + e   
/___,         
n = 1

$$\sum_{n=1}^{\infty} \frac{e^{n}}{e^{2 n} + 1}$$

Sum(exp(n)/(1 + exp(2*n)), (n, 1, oo))

Numerical answer [src]

0.535560664983911584785786496342

0.535560664983911584785786496342

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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Identical expressions

Similar expressions

Sum of series e^n/(1+e^(2*n))

The solution

Examples of finding the sum of a series