Binomial parallel resistors.

We have $\binom{n}{j}^{-1} \; = \; \frac{j!\,(n-j)!}{n!} \; = \; \frac{\Gamma(j+1)\Gamma(n+1-j)}{n!} \; = \; (n+1)B(j+1,n+1-j) \; =\; (n+1)\int_0^1 t^j(1-j)^{n-j}\,dt$ for $0 \le j \le n$ , and hence $\sum_{j=0}^n \binom{n}{j}^{-1} \; = \; (n+1)\sum_{j=0}^n \int_0^1 t^j (1-t)^{n-j}\,dt \; = \; (n+1)\int_0^1 \frac{(1-t)^{n+1} - t^{n+1}}{1 - 2t}\,dt \; = \; \frac{n+1}{2^{n+1}}\int_0^1 \frac{(1+u)^{n+1} - (1-u)^{n+1}}{u}\,du$ putting $u = 1 - 2t$ . Since $\left|\frac{1 - (1-u)^{n+1}}{u}\right| \; = \; \left|1 + (1-u) + (1-u)^2 + \cdots + (1-u)^n\right| \; \le \; n+1$ for all $0 \le u \le 1$ , we deduce that $\lim_{n \to \infty}\frac{n+1}{2^{n+1}}\int_0^1 \frac{1-(1-u)^{n+1}}{u}\,du \; = \; 0$ Since $\left| \frac{(1+u)^{n+1} - 1}{u}\right| \; = \; \left| 1 + (1+u) + (1+u)^2 + \cdots + (1+u)^n\right| \; \le \; (n+1)\big(\tfrac32)^n$ for all $0 \le u \le \tfrac12$ , we deduce that $\lim_{n \to \infty}\frac{n+1}{2^{n+1}}\int_0^{\frac12} \frac{(1+u)^{n+1}-1}{u}\,du \; = \; 0$ Thus $\lim_{n \to \infty}\sum_{j=0}^n \binom{n}{j}^{-1} \; = \; \lim_{n \to \infty} \frac{n+1}{2^{n+1}}\int_{\frac12}^1\frac{(1+u)^{n+1} - 1}{u}\,du \; = \; \lim_{n \to \infty} \frac{n+1}{2^{n+1}}\int_{\frac12}^1\frac{(1+u)^{n+1}}{u}\,du \; = \; \lim_{n \to \infty}2(n+1)\int_0^{\ln\frac43}\frac{e^{-(n+2)v}}{2e^{-v}-1}\,dv$ after the variable change $u+1 = 2e^{-v}$ . Watson's Lemma can now be used to deduce that $\lim_{n \to \infty} \sum_{j=0}^n \binom{n}{j}^{-1} \; = \; \lim_{n \to \infty} \frac{2(n+1)}{n+2} \; = \; 2$ Thus the limit as $n \to \infty$ of the effective resistance is $\tfrac12$ , making the answer $\boxed{5}$ .

We know that the value of the given expression is $\dfrac{2^{i+1}}{(i+1)(\dfrac{2^1}{1}+\dfrac{2^2}{2}+\dfrac{2^3}{3}+...+\dfrac{2^{i+1}}{i+1})}$ . When $i\rightarrow \infty$ , the result approaches the limit $\dfrac{1}{2}=\dfrac{5}{10}$ . Hence the required answer is $\boxed 5$ .

Here is a hand-waving explanation. The histogram of a row's binomial values approximates a Gaussian curve. Therefore, it is sufficient to to go far enough to see where that limit is.

When solving problems, it is a good idea to have an idea where the solution is. This is such an approximation.