# How many methods to this limit $\lim_{n\to\infty}\left(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n-1}{2^n}\right)$

find the limit $$\lim_{n\to\infty}\left(\dfrac{1}{2}+\dfrac{3}{2^2}+\cdots+\dfrac{2n-1}{2^n}\right)$$

My try: let

$$S=\dfrac{1}{2}+\dfrac{3}{2^2}+\cdots+\dfrac{2n-1}{2^n}$$ $$\dfrac{1}{2}S=0+\dfrac{1}{2^2}+\cdots+\dfrac{2n-3}{2^n}+\dfrac{2n-1}{2^{n+1}}$$ so $$\dfrac{1}{2}S=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\cdots+\dfrac{2}{2^n}-\dfrac{2n-1}{2^{n+1}}$$ so $$S=3-\dfrac{4}{2^n}-2\cdot\dfrac{n}{2^n}+\dfrac{1}{2^n}\longrightarrow 3,n\longrightarrow 3$$

solution 2:

let $$S=\sum_{n=1}^{\infty}nx^n,|x|<1|$$ $$S=x\sum_{n=1}^{\infty}nx^{n-1}=x\left(\sum_{n=1}^{\infty}x^n\right)'=x\left(\dfrac{x}{1-x}\right)'$$

Have other nice methods?Thank you

• Write each term ${2n-1\over 2^n}=\underbrace{{1\over 2^n}+\cdots+{1\over2^n}}_{(2n-1)-\rm terms}$ as the $n$th column of an infinite matrix (as in Jonas Meyer's solution here). Then compute the sum of each row of the matrix. Then compute the sum of the row sums. – David Mitra Oct 17 '13 at 9:54

## 1 Answer

If you are looking to collect other (truly) exotic solutions to this admittedly simple sum I can offer one using harmonic sums and Mellin transforms. This might well qualify as a winner where eclecticism is concerned.

Introduce $$T(x) = \sum_{n\ge 1}\frac{2n-1}{2^{xn}} = \frac{1}{\sqrt{2}^x} \sum_{n\ge 1}\frac{2n-1}{2^{xn-x\times1/2}} = \frac{1}{\sqrt{2}^x} \frac{1}{x} \sum_{n\ge 1}\frac{x(2n-1)}{\sqrt{2}^{x(2n-1)}}.$$ We will evaluate $T(x)$ by inverting the Mellin transform of $$S(x) = \sum_{n\ge 1}\frac{x(2n-1)}{\sqrt{2}^{x(2n-1)}}$$ and then put $x=1$ to get the value of the sum.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = 2k-1 \quad \text{and} \quad g(x) = x\sqrt{2}^{-x}.$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty x\sqrt{2}^{-x} x^{s-1} dx = \int_0^\infty \sqrt{2}^{-x} x^{(s+1)-1} dx = \frac{1}{(1/2\log 2)^{s+1}}\Gamma(s+1).$$ Furthermore, $$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{(2k-1)^s} = (1-2^{-s})\zeta(s).$$ Therefore the Mellin transform $Q(s)$ of $S(x)$ is given by $$Q(s) = \frac{1}{(1/2 \log 2)^{s+1}} (1-2^{-s}) \Gamma(s+1) \zeta(s).$$ The appropriate Mellin inversion integral is then given by $$\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds.$$ The pole at $s=1$ from the zeta function term has residue $$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{1}{x} \frac{1}{(1/2 \log 2)^2} (1-2^{-1})\Gamma(1) = \frac{2}{x(\log 2)^2}.$$ The remaining residues of the poles at the negative integers are given by $$- \sum_{q\ge 1} (1/2 \log 2)^{q-1} (1-2^q) \frac{(-1)^{q-1}}{(q-1)!} \frac{B_{q+1}}{q+1} x^q.$$ The trivial zeros of the zeta function term at the negative even integers cancel the poles of the gamma function term at those values. We may still let $q$ run through all negative integers because the corresponding Bernoulli numbers are zero.

Re-write this as $$- x \sum_{q\ge 1} (1/2 \log 2)^{q-1} (1-2^q) (-1)^{q-1} \frac{B_{q+1}}{(q+1)!} q x^{q-1}.$$

Now the exponential generating function of the Bernoulli numbers is given by $$\frac{t}{e^t-1} = \sum_{m\ge 0} B_m \frac{t^m}{m!}.$$ Therefore $$-1 + \frac{t}{e^t-1} = t \sum_{m\ge 1} B_m \frac{t^{m-1}}{m!}$$ and $$\left(-\frac{1}{t} + \frac{1}{e^t-1}\right)' = \sum_{m\ge 2} \frac{B_m}{m!} (m-1) t^{m-2} = \sum_{m\ge 1} \frac{B_{m+1}}{(m+1)!} m t^{m-1}.$$ This gives $$\frac{1}{t^2} - \frac{e^t}{(e^t-1)^2} = \sum_{m\ge 1} \frac{B_{m+1}}{(m+1)!} m t^{m-1}.$$ We now substitute the appropriate values of $t$ from the sum of the residues into this formula, getting two terms, the first of which gives $t=-1/2\log 2 \times x$, producing $$-x \left(\frac{1}{x^2 (1/2\log 2)^2} - \frac{1/\sqrt{2}^x}{(1/\sqrt{2}^x-1)^2}\right).$$ The second term has $t=-\log 2 \times x$ and gives $$-2x \left(\frac{1}{x^2 (\log 2)^2} - \frac{1/2^x}{(1/2^x-1)^2}\right).$$ Collecting the contributions from all residues we finally obtain $$\frac{2}{x(\log 2)^2} - \frac{1}{x (1/2\log 2)^2} + x\frac{1/\sqrt{2}^x}{(1/\sqrt{2}^x-1)^2} + \frac{2}{x (\log 2)^2} - 2x\frac{1/2^x}{(1/2^x-1)^2}.$$ Carrying out the cancellation we have that $$S(x) = x\frac{1/\sqrt{2}^x}{(1/\sqrt{2}^x-1)^2} - 2x\frac{1/2^x}{(1/2^x-1)^2}.$$ This implies that $$T(x) = \frac{1/2^x}{(1/\sqrt{2}^x-1)^2} - \frac{2}{\sqrt{2}^x} \frac{1/2^x}{(1/2^x-1)^2}.$$ In particular we have for $x=1$ that $$T(1) = \frac{1/2}{(1/\sqrt{2}-1)^2} - \sqrt{2}\frac{1/2}{(1/2-1)^2} = \frac{1/2 ((1/\sqrt{2}+1)^2)}{(1/2-1)^2} - 2\sqrt{2} \\ = 2 ((1/\sqrt{2}+1)^2) -2\sqrt{2} = 2 (3/2+2/\sqrt{2}) -2\sqrt{2} = 3.$$