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.


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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