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