Finding Discontinuities & Decreasing Intervals of a Sequence

Bashyboy
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Homework Statement


\sum_{n=1}^{\infty} \frac{n^{k-1}}{n^k+c}, where k is a positive integer.

Homework Equations


The Attempt at a Solution


I found that it was discontinuous at x = (-c)^{1/k}; and to determine if the sequence is decreasing, I took the

derivative which is--I think--f'(x) = \frac{(k-1)x^{k-2}(x^k+c)-x^{k-1}(kx^{k-1}}{(x^k+c)^2}
I am not quite sure how to simplify this, nor am I certain on how to find the intervals which the sequence is decreasing.
 
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Bashyboy said:

Homework Statement


\sum_{n=1}^{\infty} \frac{n^{k-1}}{n^k+c}, where k is a positive integer.


Homework Equations





The Attempt at a Solution


I found that it was discontinuous at x = (-c)^{1/k}; and to determine if the sequence is decreasing, I took the

derivative which is--I think--f'(x) = \frac{(k-1)x^{k-2}(x^k+c)-x^{k-1}(kx^{k-1}}{(x^k+c)^2}
I am not quite sure how to simplify this, nor am I certain on how to find the intervals which the sequence is decreasing.


Now that you've taken the derivative of f, ask yourself, what are the critical points? Those will allow you to find if f is decreasing as x → ∞.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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