Solution
- To begin, let us write the series in a general form:
[tex]\begin{gathered} \frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+... \\ \\ \text{ Factorize }\frac{1}{5}\text{ out} \\ \\ \frac{1}{5}(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...) \\ \\ \frac{1}{5}(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...) \\ \\ \frac{1}{5}+\frac{1}{5}(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...) \\ \\ \frac{1}{5}+\frac{1}{5}\sum_{k=1}^{\infty}\frac{1}{2^k} \end{gathered}[/tex]- From the above, we get a geometric series. Thus, we can apply the Ratio test
- The ratio test theorem states that:
[tex]\begin{gathered} \text{ Let p be the limit below:} \\ p=\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| \\ \\ \text{ If p exists and is less than 1, then the series converges} \\ \text{ if p exists and is greater than 1, then, the series diverges} \end{gathered}[/tex]- Applying this theorem, we have:
[tex]\begin{gathered} a_k=\frac{1}{2^k}=2^{-k} \\ a_{k+1}=\frac{1}{2^{k+1}}=2^{-(k+1)} \\ \\ p=\lim_{k\to\infty}|\frac{2^{-(k+1)}}{2^{-k}}| \\ \\ p=\lim_{k\to\infty}|\frac{2^{-k}.2^{-1}}{2^{-k}}| \\ \\ 2^{-k}\text{ crosses out} \\ \\ \therefore p=\lim_{k\to\infty}|2^{-1}| \\ \\ p=\frac{1}{2} \\ \\ \text{ This means that:} \\ p<1,\text{ implying that the series converges} \end{gathered}[/tex]- Now that we know that the series converges, and it is a geometric series, we can apply the sum to infinity formula of a geometric series to find whether it has a sum or not.
- This is done below:
[tex]\begin{gathered} \frac{1}{5}+\frac{1}{5}\sum_{k=1}^{\infty}\frac{1}{2^k} \\ \\ \frac{1}{5}+\frac{1}{5}(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...) \\ \\ \text{ The geometric series portion of the sum has a common ratio of }\frac{1}{2}\text{ because:} \\ \frac{\frac{1}{4}}{\frac{1}{2}}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2} \\ \\ \text{ The formula for sum to infinity of a geometric sequence is:} \\ S_{\infty}=\frac{a}{1-r} \\ where, \\ a=First\text{ term} \\ r=common\text{ ratio} \\ \\ \therefore S_{\infty}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \\ \\ \therefore\frac{1}{5}+\frac{1}{5}S_{\infty}=\frac{1}{5}+\frac{1}{5}(1)=\frac{2}{5} \\ \\ \text{ Thus, the series has a sum} \end{gathered}[/tex]Final Answer
The series converges and the series has a sum
