Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 70 / 3 / c

The geometric series formula and part (b) givePut $\alpha =\log 3/\log 5$. Then $5^{\alpha }=3$, so $n^{\alpha }<3^{m+1}$ andThis is the largest possible decay exponent: at $n=5^m$, an estimate with exponent $\beta >\alpha$ would require $\frac{1}{2}\le c(3/5^{\beta }{)}^m$, whose right side tends to zero. Any smaller exponent also gives a valid weaker estimate. In fact $\frac{1}{2}{n}^{-\alpha }\le E_n(f)<\frac{3}{2}{n}^{-\alpha }$ for every $n\ge 1$.