Finding the sum $S_n$ of first n terms, check the convergence of the series
$$\sum_{n=1}^{n-1}(-1)^{n-1}\frac{2n+1}{n(n+1)}$$
$u_n=(-1)^{n-1}\frac{2n+1}{n(n+1)}=(-1)^{n-1}[\frac{1}{n}+\frac{1}{n+1}]$
then $S_n=(1/1+1/2)-(1/2+1/3)+(1/3+1/4)...$ terms cancelled out, except the first and last. But last term depends of n=even or odd.
How to show $S_n\to l$, finite number for the convergence of the given series?
On
You can consider the two cases: $$\begin{align}S_{2n-1}=&\sum_{k=1}^{2n-1} (-1)^{k-1}\left[\frac{1}{k}+\frac1{k+1}\right]=\left[1+\frac12\right]-\left[\frac12+\frac13\right]+\cdots +\left[\frac{1}{2n-1}+\frac{1}{2n}\right]=\\ =&1-\frac{1}{2n};\\ S_{2n}=&\sum_{k=1}^{2n} (-1)^{k-1}\left[\frac{1}{k}+\frac1{k+1}\right]=\left[1+\frac12\right]-\left[\frac12+\frac13\right]+\cdots -\left[\frac{1}{2n}+\frac{1}{2n+1}\right]=\\ =&1-\frac{1}{2n+1};\\ \end{align}.$$ Hence: $$\lim_{n\to\infty} S_{2n-1}=\lim_{n\to\infty}S_{2n}=1.$$
We have
\begin{align} S_n &= \sum_{i=1}^{n-1}(-1)^{i-1}\frac{2i+1}{i(i+1)}\\ &= \sum_{i=1}^{n-1}(-1)^{i-1}\left[\frac1i+\frac1{i+1}\right]\\ &= \sum_{i=1}^{n-1} (-1)^{i-1}\frac1i + \sum_{i=1}^{n-1} (-1)^{i-1}\frac1{i+1}\\ &= -\sum_{i=1}^{n-1} \frac{(-1)^{i}}i + \sum_{i=2}^{n} \frac{(-1)^{i}}i\\ &= 1 + \frac{(-1)^{n}}n\\ &\xrightarrow{n\to\infty} 1 \end{align}