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

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

Question
Differential equations
asked 2021-02-21
Determine whether three series converges or diverges. if it converges, find its sum
\(\displaystyle{\sum_{{{n}={0}}}^{{\infty}}}\frac{{{\left(-{1}\right)}^{{n}}}}{{4}^{{n}}}\)

Answers (1)

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.
0

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