A question about series $$\sum\frac{1}{2^n} =  1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$
$$\sum \frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$
Why is the first one convergent and the second divergent? First has $$\lim_ {n\to \infty}\ \frac{1}{2^n}$$ and the second $$\lim_{n\to \infty}\ \frac{1}{n}$$ Both limis are $0$. Right? Now, why does the first series converge and the second diverge? The first one looks like "the half of the other".
 A: The fact that the terms go to zero is a necessary condition for convergence, but it is not sufficient. These two series show that it is not enough for the terms to get small in order for the series to converge, you need the partial sums to converge.
Remember that the series is really the limit of the partial sums:
$$\sum_{i=1}^{\infty}a_i = \lim_{n\to\infty}\left(\sum_{i=1}^n a_i\right).$$
In order for the sequence on the right to converge, the terms must get closer to one another; that's why we need (require) $a_n\to 0$ as $n\to\infty$. But this is not sufficient in order to get a sequence that converges.
The first series,
$$\sum_{i=0}^{\infty}\frac{1}{2^i}$$
converges because the sequence of partial sums converges. The $n$th partial sum is equal to
$$\sum_{i=0}^{n}\frac{1}{2^i} = \frac{1 - \frac{1}{2^{n+1}}}{\frac{1}{2}} = 2 - \frac{1}{2^n},$$
so the sequence converges to $2$ as $n\to\infty$. See also this answer.
The second series,
$$\sum_{i=1}^{\infty}\frac{1}{i}$$
does not converge because the sequence of partial sums does not converge; even though the terms differ from the preceding one by very little, they eventually get very far away from any particular term. For instance,
$$\begin{align*}
\sum_{i=1}^2\frac{1}{i} &= 1+\frac{1}{2} =\frac {3}{2}\\
\sum_{i=1}^4\frac{1}{i} &= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\
&\gt 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4} = 2\\
\sum_{i=1}^8 \frac{1}{i} &= 1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4} + \frac{1}{5}+\frac{1}{6} + \frac{1}{7}+\frac{1}{8}\\
&\gt \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + \left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)\\
&\gt 2 + \frac{1}{2}\\
\sum_{i=1}^{16}\frac{1}{i} &= \sum_{i=1}^{8}\frac{1}{i} + \sum_{i=9}^{16}\frac{1}{i}\\
&\gt \frac{5}{2} + \sum_{i=9}^{16}\frac{1}{16} = \frac{5}{2}+\frac{8}{16}\\
&= 3\\
&\vdots
\end{align*}$$
And so on. The sum up to $\frac{1}{2^n}$ will be greater than $\frac{n}{2}+1$, which gets infinitely large.
Note that the first series is not "half the other"; half the second series would be
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{6} + \frac {1}{8} + \frac{1}{10} + \frac{1}{12} + \frac{1}{14} + \frac{1}{16}+\cdots$$
you keep losing more and more terms for larger $n$ (first you are missing $\frac{1}{6}$, then $\frac{1}{10}$ and $\frac{1}{12}$ and $\frac{1}{14}$, etc.). If you did in fact have the series of "half the second", that is, if you had
$$\sum_{i=1}^{\infty}\frac{1}{2i}\quad\text{instead of}\quad \sum_{i=1}^{\infty}\frac{1}{2^i}$$
then the series would not converge, precisely because it is half a divergent series. 
