Question

Anyone know how to solve the summation
Sum[1/Sqrt[2^i], {i, 0, -1 + k}]
In other words the sum of 1/sqrt(2^i) from i=0 to k-1 ? Thanks!

Answer 1

\[\begin{align*}
\sum_{i=0}^{k-1}\frac{1}{\sqrt{2^i}}&=\frac{1}{\sqrt{2^0}}+\frac{1}{\sqrt{2^1}}+\frac{1}{\sqrt{2^2}}+\cdots+\frac{1}{\sqrt{2^{k-2}}}+\frac{1}{\sqrt{2^{k-1}}}\\
&=1+\frac{1}{\sqrt2}+\frac{1}{2}+\cdots+
\frac{1}{\sqrt{\frac{2^k}{4}}}+\frac{1}{\sqrt{\frac{2^k}{2}}}\\
&=1+\frac{1}{\sqrt2}+\frac{1}{2}+\cdots+
\frac{2}{\sqrt{2^k}}+\frac{\sqrt2}{\sqrt{2^k}}\\
&=\frac{1}{\sqrt{2^k}}\left(\sqrt{2^k}+\frac{\sqrt{2^k}}{\sqrt2}+\frac{\sqrt{2^k}}{2}+\cdots+2+\sqrt2\right)
\end{align*} \]
Hmm, I'm not sure what to do now, but according to WolframAlpha, I seem to be headed in the right direction with factoring out that 1/√(2^k):
http://www.wolframalpha.com/input/?i=sum+1%2Fsqrt%282%5Ei%29+from+i%3D0+to+i%3Dk-1

Answer 2

Thanks for the help!