What is the sum of the following infinite series

Question

What is the sum of the following infinite series: 1+(1/2)+(1/3)+(1/4) and so and until forever?

Answer 1

$\sum_{n=1}^\infty\frac{1}{n}$ is called the Harmonic Series and will never converge.

Proof:

Assume the harmonic series converges to $S$.

That is, $S= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...$

Then $S\geq 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{8}+...=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...=\frac{1}{2}+S$, which is a contradiction because $\frac{1}{2}+S$ can not be less than $S$.

Answer 2

This is not a convergent sequence so it is not possible to find the sum.

Let $\displaystyle S=\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots= (1)+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\dots+\frac{1}{7})+\dots+(\frac{1}{2^k}+\frac{1}{2^k+1}+\dots+\frac{1}{2^{k+1}-1})+\dots\ge 1+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\dots+\frac{1}{8})+\dots+(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}}+\dots+\frac{1}{2^{k+1}})+\dots=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots+2^k\frac{1}{2^{k+1}}+\dots >1+\frac{n}{2}$

forall $n\in N$ so it cannot converge.

Answer 3

This series is known as the Harmonic Series, and diverges.