Question

# Determine if the series sum_{n=0}^infty a_n is convergent or divergent if the partial sum of the n terms of the series is given below. If the series is convergent, determine the value of the series. S_n=frac{5+8n^2}{2-7n^2}

Series
Determine if the series $$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}$$ is convergent or divergent if the partial sum of the n terms of the series is given below. If the series is convergent, determine the value of the series.
$$\displaystyle{S}_{{n}}={\frac{{{5}+{8}{n}^{{2}}}}{{{2}-{7}{n}^{{2}}}}}$$

2020-11-24
Result on convergence of series:
If the sequence of partial sum $$\displaystyle{S}_{{n}}={a}_{{1}}+{a}_{{2}}+\ldots+{a}_{{n}}$$ of given series $$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}$$ is convergent sequence them the series is also convergent .
Moreover, the sum of the series is given by
$$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}=\lim_{{{n}\to\infty}}{S}_{{n}}$$
Now, the given sequence of partial sum is
$$\displaystyle{S}_{{n}}={\frac{{{5}+{8}{n}^{{2}}}}{{{2}-{7}{n}^{{2}}}}}$$
Writing the first few terms of the sequence,
$$\displaystyle{<}{\frac{{{5}+{8}}}{{{2}-{7}}}},{\frac{{{5}+{8}\times{2}^{{2}}}}{{{2}-{7}\times{2}^{{2}}}}},{\frac{{{5}+{8}\times{2}^{{3}}}}{{{2}-{7}\times{2}^{{3}}}}},\ldots{>}$$
$$\displaystyle{<}-{2.6},-{1.42},-{1.26},\ldots{>}$$
Clearly, the sequence is monotonically increasing.
Also it is bounded above by'0' that is $$\displaystyle{S}_{{n}}{<}{0},\forall{n}$$</span>,
Hence, the given sequence of partial sum is convergent.
In order to get the sum of the series, we will find the limit of the given sequence of partial sum.
$$\displaystyle\lim_{{{n}\to\infty}}{S}_{{n}}=\lim_{{{n}\to\infty}}{\frac{{{5}+{8}{n}^{{2}}}}{{{2}-{7}{n}^{{2}}}}}$$
$$\displaystyle\lim_{{{n}\to\infty}}{S}_{{n}}=\lim_{{{n}\to\infty}}{\frac{{{16}{n}}}{{-{14}{n}}}}$$
$$\displaystyle=-{\frac{{{8}}}{{{7}}}}$$
Hence,
$$\displaystyle{\sum_{{{n}={0}}}^{\infty}}{a}_{{n}}=\lim_{{{n}\to\infty}}{S}_{{n}}=\lim_{{{n}\to\infty}}{\frac{{{5}+{8}{n}^{{2}}}}{{{2}-{7}{n}^{{2}}}}}=-{\frac{{{8}}}{{{7}}}}$$