<s>[INST] 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
$$\frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 8 + \dots + \frac 1 {2n}$$
 [/INST] 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
$$\sum\limits_{n = 1}^m {\frac{1}{n}}  \sim \log \left( {m + 1} \right) + \gamma $$
</s>