Question

Determine whether the given series is convergent or divergent. Explain your answer. If the series is convergent, find its sum.
\(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}\)

Answer 1

Given the series:
\(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}\)
Rewriting the given series as sum of two series:
\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\left(\frac{3^n}{4^n}+\frac{2^{n+1}}{4^n}\right)\)
\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\frac{3^n}{4^n}+\sum_{n=0}^\infty\frac{2^{n+1}}{4^n}\)
\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty\frac{2\cdot2^n}{4^n}\)
\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{2}{4})^n\)
\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{1}{2})^n\)
Hence we get sum of two geometric series:
\(r_1=\frac34\)
\(r_2=\frac12\)
Since both \(|r_1|<1,|r_2|<1\)</span>
Both the individual geometric series converge.
The given series is sum of two converging, hence the given series \(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}\) converges.
The sum of an infinite geometric series is given as:
\(S=\frac{\text{1st Term}}{1-r}\)
The sum of the series will be:
\(\sum_{n=0}^\infty(\frac34)^n=\frac{1}{1-\frac34}\)
\(\Rightarrow\sum_{n=0}^\infty(\frac34)^n=\frac{1}{\frac14}=4\)
\(\sum_{n=0}^\infty2(\frac12)^n=\frac{2}{\frac12}=4\)
The sum of the given series will be:
\(\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=4+4\)
\(\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=8\)
Final Answer:
The given series is converging.
The sum of series is 8