Question

find the first four partial sums
k=1sigma tends to infinity (2/(5^k)-1)

Answer 1

is this
\[\frac{2}{5^k-1}\]

Answer 2

yes yes..

Answer 3

replace \(k\) by 1 get
\[\frac{2}{5^1-1}=\frac{2}{4}=\frac{1}{2}\]so that is your first partial sum

Answer 4

replace \(k\) by 2 get
\[\frac{2}{5^2-1}=\frac{1}{12}\] so your second partial is
\[\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\]

Answer 5

sory sory 5^(k-1)

Answer 6

oh ok

Answer 7

replace \(k\) by 1 get
\[\frac{2}{5^0}=2\] so that is the first partial

Answer 8

now replace \(k\) by 2 and get
\[\frac{2}{5^1}=\frac{2}{5}\] second partial is
\[2+\frac{2}{5}=\frac{12}{5}\]

Answer 9

etc etc

Answer 10

i.e. replace k by successive integers, and add

Answer 11

thnks. how can i find the series closed form of nth partial sums and how can i find this series in converge or diverge

Answer 12

plz help me.;(