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Test the following series for convergence \(\sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^{n-1}\)

\(\sum_{n=1}^{\infty} a r^{n-1}\)

Gemetric series (G.s) with \(r=\frac{1}{2}<1\)

convergent series 

\(a=3 \cdot\left(\frac{1}{2}\right)^{1-1}=3\left(\frac{1}{2}\right)^{0}=3(1)=3\)

sum \(=\frac{a}{1-r}=\frac{3}{1-\frac{1}{2}}=6\)

Is the following series convergent? \(\sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^{n}\)

Geometric series with \(r=\frac{1}{2}<1\)

Convergent series

\(a=3\left(\frac{1}{2}\right)^{1}=\frac{3}{2}\)

Sum \(=\frac{a}{1-r}=\frac{\frac{3}{2}}{1-\frac{1}{2}}=\frac{\frac{3}{2}}{\frac{1}{2}}=3\)

Is the following series convergent? \(\sum_{n=2}^{\infty} 3\left(\frac{1}{2}\right)^{n+1}\)

Gometric series (G.S) with \(r=\frac{1}{2}<1\)

Convergent series

\(a=3\left(\frac{1}{2}\right)^{2+1}=3\left(\frac{1}{2}\right)^{3}=3\left(\frac{1}{8}\right)=\frac{3}{8}\)

sum \(=\frac{a}{1-r}=\frac{\frac{3}{8}}{1-\frac{1}{2}}=\frac{\frac{3}{8}}{\frac{1}{2}}=\frac{3}{8} \times \frac{2}{1}=\frac{3}{4}\)

Is the following series convergent? \(\sum_{n=0}^{\infty}(-\ln (2))^{n}\)

Gometric series with \(|r|=|-\ln (2)|=\ln (2)=0.69 \ldots<1\)

convergent series 

\(a=(-\ln (2))^{0}=1\)

sum \(=\frac{a}{1-r}=\frac{1}{1-(-\ln (2))}=\frac{1}{1+\ln (2)}\)

Is the following series convergent? \(\sum_{n=0}^{\infty}\left[\cos \left(\frac{\pi}{3}\right)\right]^{n}\)

Geometric series with \(|r|=\left|\cos \frac{\pi}{3}\right|=\frac{1}{2}<1\)

Convergent series

\(a=\left[\cos \frac{\pi}{3}\right]^{0}=1\)

sum \(=\frac{a}{1-r}=\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2\)

\(\sum_{n=0}^{\infty}\left[\cos \frac{\pi}{3}\right]^{n}=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n}\)

Find the value of t such that \(\sum_{n=1}^{\infty}\left(\frac{t}{t+2}\right)^{n}=3\)

Geometric series (G.S) \(=\sum_{n=1}^{\infty}\left(\frac{t}{t+2}\right)^{n}\) with \(r=\frac{t}{t+2}\)

\(a=\left(\frac{t}{t+2}\right)^{1}=\frac{t}{t+2}\)

Sum \(=\frac{a}{1-r}=\frac{\frac{t}{t+2}}{1-\frac{t}{t+2}}\)

\(=\frac{\frac{t}{t+2}}{\frac{t+2}{t+2}-\frac{t}{t+2}}=\frac{\frac{t}{t+2}}{\frac{t+2-t}{t+2}}\)

Sum \(=\frac{t}{2}=3 \quad \Rightarrow t=3 \cdot 2=6\)

\(t=6\)

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