Because
$$1+1+1+1+1+...=\sum_{n=0}^{\infty} 1$$
doesn't converge. This sum is not a finite number. So if there is any hope of an infinite series converging, the terms MUST go to zero. It is not true the other way. Terms going to zero is not enough for convergence with the divergent harmonic series being the classic example.
Addendum:
BDillan, you seem confused about two different things.
1.First you seem confused about logic and what the contrapositive statement is for a given statement. If the original statement is $p \Rightarrow q$ then the contrapositive is $\sim q \Rightarrow \sim p$. Your statement 1 is
If $\sum a_k$ converges, then $\displaystyle\lim_{k\to\infty}a_k = 0$.
You statement 2 is
If $\sum a_k$ diverges, then $\displaystyle\lim_{k\to\infty} a_k \neq 0$.
You statement 1 is true. Your statement 2 is not the contrapositive of statement 1 therefore statement 1 and statement 2 are not equivalent. Statement 2 is the inverse of statement 1 ($\sim p \Rightarrow \sim q$). The correct contrapositive is
If $\displaystyle\lim_{k\to\infty} a_k \neq 0$ then $\sum a_k$ diverges.
2.Since the truth of a statement has nothing to do with the truth of its inverse, in this case your statement 2 is false. You seem to confusing the terms of the sequence
$$\{a_n\}_{n=1}^{\infty}=\{a_1,a_2,a_3,...\}$$
and the sequence of partial sums
$$S_1 = a_1$$$$S_2 = a_1+a_2$$$$S_3 = a_1+a_2+a_3...$$$$S_n = \sum_{i=1}^{n} a_i...$$
An infinite sum converges if an only if the limit of partial sums exists and is finite because
$$\sum_{i=1}^{\infty} a_i = \displaystyle\lim_{n\rightarrow\infty}S_n.$$
So once again, if the sum is convergent, then the terms must be going to zero (original statement). If the terms don't go to zero then sum is not convergent (the contrapositive). If the terms do go to zero then we have no idea what happens (the inverse) and we need further investigation.
- The series can converge $\sum \frac{1}{n^2}.$
- It can diverge to positive infinity $\sum \frac{1}{n}.$
- It can diverge to negative infinity $\sum \frac{-1}{n}.$