# What kind of series is this, and how do I sum it?

$\displaystyle\sum{\frac1{a_n}}$, e.g. $$\frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \dots + \frac 1 n$$ or
$$\frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 8 + \dots + \frac 1 {2n}$$

You are looking for harmonic numbers. An excellent article can be found here. The corresponding (diverging) series is the harmonic series. To sum it, use the integral representation given in the first article:

$$\int_0^1 \frac{1 - x^n}{1 - x}\,{\rm d}x = \int_0^1 (1 + x + \cdots + x^{n-1})\ {\rm d}x = 1 + \frac12 + \cdots + \frac1{n} = H_n$$

Maybe if it's not clear the left one you mentioned is $H_n$, the right one is obviously $\frac12\cdot H_n$.

That is a harmonic series. You can approximate it by integration, as explained in the link. Note that if $n \to \infty$, the sum diverges.

A nice approximation for $n \to \infty$ can be derived from

\eqalign{ & \log \left( {1 + x} \right) = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{{{x^{k + 1}}}}{{k + 1}}} = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - + \cdots \cr & \log \left( {1 + \frac{1}{n}} \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{n^{k + 1}}}}\frac{1}{{k + 1}}} = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - + \cdots \cr}

$$\log \left( {n + 1} \right) - \log n = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^k}\frac{1}{{{n^{k + 1}}}}\frac{1}{{k + 1}}} = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - + \cdots$$

If we sum for $n=1,2,3,\dots,m$ we have

$$\log \left( {m + 1} \right) = \sum\limits_{n = 1}^m {\frac{1}{n}} - \frac{1}{2}\sum\limits_{n = 1}^m {\frac{1}{{{n^2}}}} + \frac{1}{3}\sum\limits_{n = 1}^m {\frac{1}{{{n^3}}}} - + \cdots$$ or

$$\sum\limits_{n = 1}^m {\frac{1}{n}} - \log \left( {m + 1} \right) = \frac{1}{2}\sum\limits_{n = 1}^m {\frac{1}{{{n^2}}}} - \frac{1}{3}\sum\limits_{n = 1}^m {\frac{1}{{{n^3}}}} + - \cdots$$

If we let $m\to \infty$ , we have that

$$\mathop {\lim }\limits_{m \to \infty } \left[ {\sum\limits_{n = 1}^m {\frac{1}{n}} - \log \left( {m + 1} \right)} \right] = \frac{1}{2}\zeta \left( 2 \right) - \frac{1}{3}\zeta \left( 3 \right) + \frac{1}{4}\zeta \left( 4 \right) - + = \sum\limits_{n = 2}^\infty {{{\left( { - 1} \right)}^n}\frac{{\zeta \left( n \right)}}{n}}$$

where $$\zeta \left( n \right) = \sum\limits_{k = 1}^\infty {\frac{1}{{{k^n}}}}$$

The value of the right hand side is know as Euler's constant, $\gamma \approx 0.577$, and this result tells us that for large values of $m$, we can approximate the $m$th harmonic number by

$$\sum\limits_{n = 1}^m {\frac{1}{n}} \sim \log \left( {m + 1} \right) + \gamma$$