Determine whether three series converges or diverges. if it converges, find its sum sum_(n=0)^(oo)((-1)^n)/4^n

Question
Differential equations
Determine whether three series converges or diverges. if it converges, find its sum
$$\displaystyle{\sum_{{{n}={0}}}^{{\infty}}}\frac{{{\left(-{1}\right)}^{{n}}}}{{4}^{{n}}}$$

2021-02-22
1)Here the series is given by $$\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}{\left(-{1}\right)}^{{n}}{a}_{{n}}={\sum_{{{n}={0}}}^{{\infty}}}\frac{{{\left(-{1}\right)}^{{n}}}}{{4}^{{n}}}$$.This is a alternating series with $$\displaystyle{a}_{{n}}=\frac{{1}}{{4}^{{n}}}.{C}\le{a}{r}{l}{y}{a}_{{n}}$$ is a decreasing sequence and
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{a}_{{n}}=\lim_{{{n}\rightarrow\infty}}\frac{{1}}{{4}^{{n}}}={0}$$
Therefore by Leibnit'z rule the given series $$\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}\frac{{{\left(-{1}\right)}^{{n}}}}{{4}^{{n}}}$$ is convergent.
2)Consider the series $$\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}\frac{{1}}{{n}}$$.The n-th partial sum of the series is given by
$$\displaystyle{S}_{{n}}={1}+\frac{{1}}{{2}}+\frac{{1}}{{3}}+\ldots+\frac{{1}}{{n}}$$.
Now
$$\displaystyle{S}_{{{2}^{{n}}}}={1}+\frac{{1}}{{2}}+\frac{{1}}{{3}}+\frac{{1}}{{4}}+\frac{{1}}{{5}}+\frac{{1}}{{6}}+\frac{{1}}{{7}}+\frac{{1}}{{8}}+\ldots+\frac{{1}}{{2}^{{n}}}$$
$$\displaystyle={1}+\frac{{1}}{{2}}+{\left(\frac{{1}}{{3}}+\frac{{1}}{{4}}\right)}+{\left(\frac{{1}}{{5}}+\frac{{1}}{{6}}+\frac{{1}}{{7}}+\frac{{1}}{{8}}\right)}+\ldots+\frac{{1}}{{2}^{{n}}}$$
$$\displaystyle\Rightarrow{1}+\frac{{1}}{{2}}+{\left(\frac{{1}}{{4}}+\frac{{1}}{{4}}\right)}+{\left(\frac{{1}}{{8}}+\frac{{1}}{{8}}+\frac{{1}}{{8}}+\frac{{1}}{{8}}\right)}+\ldots+{\left(\frac{{1}}{{n}}+\ldots+{2}^{{{n}-{1}}}\frac{{1}}{{2}^{{n}}}\right)}$$
$$\displaystyle={1}+\frac{{1}}{{2}}+\frac{{1}}{{2}}+\frac{{1}}{{2}}+\ldots+\frac{{1}}{{2}}$$
$$\displaystyle={1}+\frac{{n}}{{2}}$$
but $$\displaystyle{\left\lbrace\frac{{n}}{{2}}\right\rbrace}$$ is a divergent sequence. This shows that $$\displaystyle{S}_{{{2}^{{n}}}}$$
is also divergent and hence the series $$\displaystyle{\sum_{{{n}={1}}}^{{\infty}}}\frac{{1}}{{n}}$$ is divergent.

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