Question

find a closed form of nth partial sums
and determined whether the series in converges
k=1sigma k tends to infinity (2/(5^(k-1))

Answer 1

plz help me..and one help me.:(

Answer 2

any one there..

Answer 3

firstly closed form of the serie
\[\sum_{k=1}^{n} \frac{2}{5^{k-1}}=2 \sum_{k=1}^{n} \frac{1}{5^{k-1}}=2(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{n-1}})\]
thats a geometric serie

Answer 4

do u know what is the closed form of geometric serie?

Answer 5

no brother i dont know i saw first time tthis series

Answer 6

what is the closed form of nth partial sums ? can u explain..?

Answer 7

well for geometric serie like
a+aq+aq^2+aq^3+...aq^(n-1)

closed form is
\[\frac{a(1-q^n)}{1-q}\]

Answer 8

closed form of nth partial sums is the partial sum of all the terms from the first and up to the nth which is aq^(n-1)

Answer 9

in my question what is the cloased form of nth partial sum>?

Answer 10

ok u have a=1 and q=1/5 then closed form of nth partial sums is
\[\frac{1-(\frac{1}{5})^n}{1-\frac{1}{5}}\]

Answer 11

i forgot the constant coefficient 2

Answer 12

actually u will have
\[S _{n}=2\frac{1-(1/5)^n}{1-1/5}\]

Answer 13

whether the series is converges..?

Answer 14

well for that case u have to calculate
\[\lim_{n \rightarrow \infty} S _{n}\]
We will say that series is convergent if and only if the lim is exist

Answer 15

now u have
\[\lim_{n \rightarrow \infty} 2\frac{1-(1/5)^n}{1-1/5}=\frac{2}{1-1/5}=\frac{10}{4}=2.5\]

Answer 16

then the series is converges

Answer 17

(1/5)^ infinity is zero..?

Answer 18

yes thats right

Answer 19

brother where is series converges or diverges

Answer 20

brother in this question firsr four partial sum is 310/125 my ans is right or wrong.?

Answer 21

well thats right

Answer 22

brother any direct formula u know tho find series converges or diverges.?

Answer 23

no... any direct formula

Answer 24

bhahi jaza ALLAH.. allah give u many happiness thanks

Answer 25

welcome