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Rewriting the recursion yields $$1 = \sum\limits_{1\leq i \leq k} \frac{i\ f(i)}{k+i-1}.$$ By assuming $f(k)=\frac{1}{k \ln 2} (1+o(1))$, 1+o(1))$--- where the little o-Notation$o(1)$simply means that is it is smaller than every constant for large$k$, k$ --- and inserting this, we get $$\sum\limits_{1\leq i \leq k} \frac{i\ f(i)}{k+i-1} = \frac{1}{\ln 2} \sum\limits_{1\leq i \leq k} \frac{1+o(1)}{k+i-1} = \frac{1+o(1)}{\ln 2} \left(\ln (2k-1)-\ln(k)\right) = \frac{1+o(1)}{\ln 2} \ln\left(2-\frac1k\right) \to 1$$ for $k\to\infty$. I used $\sum_{1\leq i \leq k} \frac1i = \ln k + \gamma + O(1/n)$o(1)$. Therefore the assumtion holds asymptotically, since it fulfils the recursion. The bound$f(k)<2/k$follows from that --- at least, there is an$K$, such that for all$k\geq K$the inequality$f(k)<2/k$holds. 1 Rewriting the recursion yields $$1 = \sum\limits_{1\leq i \leq k} \frac{i\ f(i)}{k+i-1}.$$ By assuming$f(k)=\frac{1}{k \ln 2} (1+o(1))$, where the little o-Notation simply means that is is smaller than every constant for large$k$, and inserting this, we get $$\sum\limits_{1\leq i \leq k} \frac{i\ f(i)}{k+i-1} = \frac{1}{\ln 2} \sum\limits_{1\leq i \leq k} \frac{1+o(1)}{k+i-1} = \frac{1+o(1)}{\ln 2} \left(\ln (2k-1)-\ln(k)\right) = \frac{1+o(1)}{\ln 2} \ln\left(2-\frac1k\right) \to 1$$ for$k\to\infty$. I used$\sum_{1\leq i \leq k} = \ln k + \gamma + O(1/n)$. Therefore the assumtion holds asymptotically, since it fulfils the recursion. The bound$f(k)<2/k$follows from that --- at least, there is an$K$, such that for all$k\geq K$the inequality$f(k)<2/k\$ holds.