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Sum of series:
1/((2n-1)(2n+1))
17
(-1)^nx^2n+1/(2n+1)!
1/4^x
Identical expressions
one /(n(log(n))^k)
1 divide by (n( logarithm of (n)) to the power of k)
one divide by (n( logarithm of (n)) to the power of k)
1/(n(log(n))k)
1/nlognk
1/nlogn^k
1 divide by (n(log(n))^k)

Sum of series/ 1/(n(log(n))^k)

Sum of series 1/(n(log(n))^k)

The solution

You have entered [src]

  oo           
____           
\   `          
 \        1    
  \   ---------
  /        k   
 /    n*log (n)
/___,          
n = 1

$$\sum_{n=1}^{\infty} \frac{1}{n \log{\left(n \right)}^{k}}$$

Sum(1/(n*log(n)^k), (n, 1, oo))

The answer [src]

  oo          
____          
\   `         
 \       -k   
  \   log  (n)
  /   --------
 /       n    
/___,         
n = 1

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

Sum(log(n)^(-k)/n, (n, 1, oo))

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
```