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OpenStudy (anonymous):
k=1 sigma k tends to infinity ((1/2^(k))-(1/2^(k-1)))
how can calculte the sum of this serires.
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OpenStudy (anonymous):
Is the is what you are asking for
\[
\sum_{k=1}^\infty 2^{-k}-2^{1-k}
\]
OpenStudy (anonymous):
1/2^k
OpenStudy (anonymous):
1/2^(k-1)
OpenStudy (anonymous):
You are probably more familiar with this
\[
\sum_{k=1}^\infty \left(\frac {1}{2^k} - \frac {1}{2^{k-1}}\right)
\]
OpenStudy (anonymous):
yes this is right.
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OpenStudy (anonymous):
sorry sorry 1/2^(K+1)
OpenStudy (anonymous):
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.
OpenStudy (anonymous):
\[
\sum_{k=1}^\infty \left(\frac {1}{2^k} - \frac {1}{2^{k+1}}\right)
\]
Is this final now.
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
\[
\frac {1}{2^1} - \frac {1}{2^{2}}
\\
\frac {1}{2^2} - \frac {1}{2^{3}}
\\
\frac {1}{2^3} - \frac {1}{2^{4}}
\\
\cdots
\frac {1}{2^k} - \frac {1}{2^{k+1}}
\\
s_k=\frac {1}{2 } - \frac {1}{2^{k+1}}
\\
\]
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OpenStudy (anonymous):
\[
\lim_{k->\infty} s_k =\frac 1 2
\]
OpenStudy (anonymous):
Did you understand what I did?
OpenStudy (anonymous):
no dear..
OpenStudy (anonymous):
Ok, think about it some more.
OpenStudy (anonymous):
bro can explain now..?
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