Proof for a Statement on Series In my textbook after this theorem comes another statement(I wasn't sure to call it a theorem). The proof for the theorem is given but I don't know why the statement is true. Can you help me prove it?
Theorem If $\Sigma^{\infty}_{n=1}a_n$ converges, then $\lim_{n\to\infty}a_n = 0$. Therefore, if $\lim_{n\to\infty}a_n$ does not exist, or exists but is not zero, then the series $\Sigma^{\infty}_{n=1}a_n$ is divergent.
Statement If the sequence of terms $\{a_n\}$ tends to a non-zero limit $L$, then $\Sigma^{\infty}_{n = 1}a_n$ diverges to infinity if $L > 0$ and diverges to negative infinity if $L < 0$.

Think about it this way: if your sequence $a_n$ converges to a positive number $L>0$, then the series, which is the sum of the infinite terms of the sequence, will come to a point where it will basically be summing $L$ an infinite number of times. This can only result in the series diverging to $+\infty$.
On the other hand, if $L<0$, you are adding a negative value an infinite number of times. This can only result in the series diverging to $-\infty$.
The only way that a series can converge is if it comes to a point where the infinite terms you are adding are basically equal to zero. This requires $L=0$.
