shobber wrote: Given: \[ S = 1 + \frac{1}{1 + \frac{1}{3}} + \frac{1}{1 + \frac{1}{3} + \frac{1} {6}} + \cdots + \frac{1}{1 + \frac{1}{3} + \frac{1}{6} + \cdots + \frac{1} {1993006}} \] where the denominators contain partial sums of the sequence of reciprocals of triangular numbers (i.e. $k=\frac{n(n+1)}{2}$ for $n = 1$, $2$, $\ldots$,$1996$). Prove that $S>1001$. We have \[ \frac{1}{k_n}=\frac{2}{n(n+1)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right) \] Thus \[ \sum\limits_{n=1}^m\frac{1}{k_n}=\sum\limits_{n=1}^m2\left(\frac{1}{n}-\frac{1}{n+1}\right)=2\left(1-\frac{1}{m+1}\right)=2\frac{m}{m+1} \\\Rightarrow S=\sum\limits_{m=1}^{1996}\frac{1}{\sum\limits_{n=1}^m\frac{1}{k_n}} =\sum\limits_{m=1}^{1996}\frac{m+1}{2m}=\frac{1}{2}\sum\limits_{m=1}^{1996}\left(1+\frac{1}{m}\right)=998+\frac{1}{2}\sum\limits_{m=1}^{1996}\frac{1}{m} \] It remains to show, that $\sum\limits_{m=1}^{1996}\frac{1}{m}>6$. I'm too lazy to show this, but it is definitely not difficult (unless it's wrong ).

Werwulff wrote: It remains to show, that $ \sum\limits_{m = 1}^{1996}\frac {1}{m} > 6$. I'm too lazy to show this, but it is definitely not difficult (unless it's wrong ). It is indeed easy. $ 1+\frac12+\frac13+\frac14+\ldots+\frac1{1996}>1+(\frac12)+(\frac13+\frac14)+(\frac15+\frac16+\frac17+\frac18)+\ldots+(\frac1{513}+\frac1{514}+\ldots+\frac1{1024})>1+(\frac12)+(\frac14+\frac14)+(\frac18+\frac18+\frac18+\frac18)+...+(\frac1{1024}+...+\frac1{1024})=1+\frac12+\frac12+\frac12+...+\frac12+\frac12=6$.

Let $a_n=1+\frac13+\frac16+\ldots+\frac2{n(n+1)}$. Note that $S=\sum_{i=1}^{1996}\frac1{a_i}$. Now we can easily calculate $a_n$ with partial fraction decomposition and telescoping series: $$a_n=\sum_{i=1}^n\frac2{i(i+1)}=\sum_{i=1}^n\left(\frac2i-\frac2{i+1}\right)=2-\frac2{n+1}=\frac{2n}{n+1}.$$Then: $$S=\sum_{i=1}^{1996}\frac1{a_i}=\sum_{i=1}^{1996}\frac{i+1}{2i}=\sum_{i=1}^{1996}\left(\frac1{2i}+\frac12\right)=\frac12\sum_{i=1}^{1996}\frac1i+998$$so it suffices to show that $\sum_{i=1}^{1996}\frac1i>6$. Since: $$1+\frac12+\ldots+\frac1{1996}>1+\frac12+\frac12+\frac14+\frac14+\frac14+\frac14+\frac18+\ldots+\frac1{1024}+\frac1{1024}=6$$we're done.