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Mathematics 13 Online
OpenStudy (anonymous):

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

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.

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):

\[ \frac {1}{2^1} - \frac {1}{2^{2}} \\ \frac {1}{2^2} - \frac {1}{2^{3}} \\ \frac {1}{2^3} - \frac {1}{2^{4}} \\ \cdots \frac {1}{2^k} - \frac {1}{2^{k+1}} \\ s_k=\frac {1}{2 } - \frac {1}{2^{k+1}} \\ \]

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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