Finding total elements in a series

Question

I have a confusion.

How many terms in the following series are needed to make a sun greater than 5/2?

1/2 + 1/3 + 1/4 + 1/5 + 1/6 + .........

Is there any shortcut technique to find this out?

Answer 1

One may observe that, since $x \mapsto \dfrac1x$ is decreasing over $[2,\infty)$, one has $$ \ln N-\ln 2=\int_2^N \frac1x\,dx\le\sum_{n=2}^N \frac1n $$ giving that $$ \frac52\leq\ln N-\ln 2 \implies 2\:e^{2.5}\approx 25\le N $$ is sufficient.

Answer 2

Observe that, for $k\geq 1$, \begin{align*} \sum_{n=2}^{2^k}\frac{1}{n}&=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots+\frac{1}{2^k}\\ &>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots+\frac{1}{2^k} \\ &=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots+\frac{1}{2} \qquad (\textrm{k times}) \\ &=\frac{k}{2}. \end{align*} That is, $$ \sum_{n=2}^{2^k}\frac{1}{n}>\frac{k}{2} $$ Using this (crude) estimate, we need $2^5=32$ terms.