<s>[INST] Help calculating a sum with exponential function 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.
 [/INST] 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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