<s>[INST] Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)}  $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if  $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)}  \right|$$ converges.
i tried to check for convergence with the 2 criterias a) decreasing and b) zero sequence but yes then i tried to transform the equotation to
$$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n(n+1)} $$
the first fraction is "ok" - the 2nd one i did partial fraction decomposition and finally got $$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n} - \frac{1}{n+1} $$
i then tried to see a pattern by find out the first sequences but im not sure if i'm on the right track.
 [/INST] i think this is an easy answer
$$ \sum_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$ 
$$ \frac{2n+3}{n(n+1)} = \frac{3(n+1) - n}{n(n+1)} = \frac{3}{n} - \frac{1}{n+1} $$
$$ \sum_{n=1}^{\infty} \left(\frac{3(-1)^n}{n} - \frac{(-1)^n}{n+1} \right) $$
$$ = \int_0^1 3\sum_{n=1}^{\infty} (-1)^n x^{n-1} - \sum_{n=1}^{\infty} (-1)^n x^n \ dx $$
$$ = \int_0^1 -\frac{3}{1+x} + \frac{x}{x+1}  \ dx $$
$$ =\int_0^1 \frac{x+1 - 4}{x+1} \ dx = \int_0^1 1 - \frac{4
}{x+1} \ dx = \left[x - 4\ln(x+1) \right]_0^1 = 1 - 4\ln 2 $$
i think you know $$ \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$
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