Question

Does the following infinite geometric series diverge or converge? Explain. 1/5+1/25+1/125+1/625

Answer 1

can somebody please help me?

Answer 2

please help

Answer 3

\[\sum_{k=1}^{\infty}\left( \frac{ 1 }{ 5 } \right)^k\]
this is a geometric series with r < 1 so it converges

Answer 4

does it have a sum

Answer 5

That's your sum: \[\frac{ 1 }{ 5 }+\frac{ 1 }{ 25 }+\frac{ 1 }{ 125 }...=\sum_{1}^{\infty}(\frac{ 1 }{ 5 })^{k}=\lim_{N \rightarrow +\infty} \sum_{1}^{N}(\frac{ 1 }{ 5 })^{k}=\lim_{N \rightarrow +\infty} \frac{ 1 }{ 5 }\frac{ 1-(\frac{ 1 }{ 5 })^{N} }{ 1-\frac{ 1 }{ 5 } }=\frac{ 1 }{ 4 }\]

Answer 6

\[\sum_{k=0}^{\infty}r^k=\frac{ 1 }{ 1-r },\,\,\forall \,\,0<r<1\]Since your series starts with k=1 you would subtract the 0th term (which is 1) from the sum of this series.
\[r=\frac{ 1 }{ 5 }\Rightarrow \sum_{k=0}^{\infty}\left( \frac{ 1 }{ 5 } \right)^k=1+\sum_{k=1}^{\infty}\left( \frac{ 1 }{ 5 } \right)^k=\frac{ 1 }{ 1-\frac{ 1 }{ 5 } }=\frac{ 5 }{ 4 }=1+\frac{ 1 }{ 4 }\]\[\Rightarrow\sum_{k=1}^{\infty}\left( \frac{ 1 }{ 5 } \right)^k=\frac{ 1 }{ 4 }\]