Question

k=1 sigma k tends to infinity ((1/2^(k))-(1/2^(k-1)))
how can calculte the sum of this serires.

Answer 1

Is the is what you are asking for
\[
\sum_{k=1}^\infty 2^{-k}-2^{1-k}
\]

Answer 2

1/2^k

Answer 3

1/2^(k-1)

Answer 4

You are probably more familiar with this
\[
\sum_{k=1}^\infty \left(\frac {1}{2^k} - \frac {1}{2^{k-1}}\right)
\]

Answer 5

yes this is right.

Answer 6

sorry sorry 1/2^(K+1)

Answer 7

This is a telescoping series. Write the first term on the first line the second then the third
and see if you sum sum them you get many cancellations.

Answer 8

\[
\sum_{k=1}^\infty \left(\frac {1}{2^k} - \frac {1}{2^{k+1}}\right)
\]
Is this final now.

Answer 9

yes

Answer 10

\[
\frac {1}{2^1} - \frac {1}{2^{2}}
\\

\frac {1}{2^2} - \frac {1}{2^{3}}
\\
\frac {1}{2^3} - \frac {1}{2^{4}}
\\

\cdots
\frac {1}{2^k} - \frac {1}{2^{k+1}}
\\

s_k=\frac {1}{2 } - \frac {1}{2^{k+1}}
\\

\]

Answer 11

\[
\lim_{k->\infty} s_k =\frac 1 2
\]

Answer 12

Did you understand what I did?

Answer 13

no dear..

Answer 14

Ok, think about it some more.

Answer 15

bro can explain now..?