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So I have a sum to calculate:

$$\sum_{n=1}^{\infty} \frac{n-2}{n2^{n-1}+2^{n+1}}$$

So I first did this:

$$\sum_{n=1}^{\infty} \frac{n-2}{2^{n-1}(n+4)}$$

So first i thought to make this into a function of some sort and use derivation and integration to make it into something easier to calculate, but am having trouble doin so.

Is this a long shot? $$ f(x)= \sum_{n=1}^{\infty} \frac{n-x}{x^{n-1}(n+x^2)}$$ Since $f(2)$ equals the original sum.

Is this the correct way to define and use it? If not any help would be appreciated, thank you in advance.

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Let's see if we can split this up:

$$\dfrac{n-2}{2^{n-1}(n+4)} = \dfrac{(n+4)-6}{2^{n-1}(n+4)} = \dfrac{1}{2^{n-1}}-\dfrac{6}{2^{n-1}(n+4)}$$

So, this sum becomes:

$$\sum_{n\ge 1} \dfrac{n-2}{2^{n-1}(n+4)} = 2-\sum_{n\ge 1} \dfrac{6}{2^{n-1}(n+4)}$$

Now, let's work with this summation:

$$\sum_{n\ge 1} \dfrac{6}{2^{n-1}(n+4)} = 192\sum_{n\ge 5} \dfrac{1}{n2^n} = 192\left(\sum_{n\ge 1} \dfrac{1}{n2^n}-\sum_{n=1}^4 \dfrac{1}{n2^n}\right)$$

This final sum is easy to calculate:

$$\sum_{n\ge 1} \dfrac{1}{n2^n} = \log 2$$

So, putting it all together, we have:

$$2-192\left(\log 2 - \dfrac{1}{2}-\dfrac{1}{8}-\dfrac{1}{24}-\dfrac{1}{64}\right) = 133-192\log 2$$

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