# Summation of the n'th root of $2^i$

So I'm basically trying to write an explicit formula for this sum here:

Originally, I was asked to use the definition of an integral to evaluate: $$\int_{0}^{1} 2^x dx$$ so I rewrote this as a reimann sum to get:

$$\lim_{n\to\infty} {\frac {1}{n}}\sum_{i=1}^{n} 2^{\frac {i}{n}}$$

How do you evaluate that summation though? I'm completely lost on that. I tried to write it as a geometric series but that's not possible. It's not a p series either. How am I supposed to sum that?

Here is the way the professor did it but no work is shown for the summation at all... • Do you know the indefinite integral of $2^x$? Note that $2^x = e^{x \ln 2}$, so maybe definition of integral could mean a use of Fundamental theorem of calculus? – Teresa Lisbon Dec 5 '16 at 5:49
• No it explicitly asked us to use reimann sums. It's even in the solutions. But no explanation was given as to how that sum was obtained. – Future Math person Dec 5 '16 at 5:51
• Then it's best you post the intermediate result, or the equality you don't understand in the answer. Surely there must be steps in the evaluation, right? – Teresa Lisbon Dec 5 '16 at 5:52
• Fine, so the summation is a standard geometric series. Do you notice that? So the formula for geometric summation can be used. Do you need me to mention it explicitly? If so, then $\sum_{i=1}^n a^i = \frac{a^{n+1}-1}{a-1} - 1$. Put in $a=2^{\frac{1}{n}}$ and simplify. – Teresa Lisbon Dec 5 '16 at 5:55
• It is a geometric series, right You are summing the powers of $2^{\frac 1n}$ from $i=1$ to $n$. It looks like a sequence where the ratio between consecutive terms is constant, right? Now the formula comes into play. – Teresa Lisbon Dec 5 '16 at 5:58

Take the Riemann summ approximation of order $$n$$ to the integral above: $$\int_0^1 2^x \approx \frac{1}{n} \sum_{k=1}^n 2^{k/n} =\frac{1}{n} \sum_{k=1}^n (2^{1/n})^k.$$ The sum on the right side is a geometric sum on $$z=2^{1/n}$$ which, as widely known, equals $$\frac{1-z^{n+1}}{1-z }=\frac{1-(2^{1/n})^{n+1}}{1-2^{1/n} }.$$ Substituted in the original equation gives $$\int_0^1 2^x = \lim_{n\to\infty} \frac{1}{n} \frac{1-(2^{1/n})^{n+1}}{1-2^{1/n} }.$$ Place $$1=\lim_{n\to\infty} 2^{1/n}$$ at the top of the fraction and the remaining part of the proof is explicit from here on in the image you shared.