Proof of the limit comparison test

Proof

1. Apply the definition of the limit with $\varepsilon = c/2 > 0$. There is $N$ such that $\abs{a_n/b_n - c} < c/2$ for all $n \ge N$.
2. So $c/2 < a_n / b_n < 3c/2$ for $n \ge N$. Since $b_n > 0$, this gives $a_n < \tfrac{3c}{2} b_n$ and $b_n < \tfrac{2}{c} a_n$ for $n \ge N$.
3. Both series have nonnegative terms. By the criterion for nonnegative series, each converges if and only if its partial sums are bounded above.
4. Assume that $\sum b_n$ converges, and let $B$ be its sum. Since $b_n \ge 0$, every finite sum of terms $b_n$ is at most $B$. This follows from the same criterion, as in the comparison test.
5. For $M \ge N$, step 2 gives $$\sum_{n=0}^{M} a_n \le \sum_{n=0}^{N-1} a_n + \frac{3c}{2} \sum_{n=N}^{M} b_n \le \sum_{n=0}^{N-1} a_n + \frac{3c}{2} B.$$
6. The right side of step 5 does not depend on $M$. For $M < N$, the partial sum is at most $\sum_{n=0}^{N-1} a_n$. So the partial sums of $\sum a_n$ are bounded above, and $\sum a_n$ converges by step 3.
7. Conversely, assume that $\sum a_n$ converges, with sum $A$. In the same way, for $M \ge N$, $$\sum_{n=0}^{M} b_n \le \sum_{n=0}^{N-1} b_n + \frac{2}{c} A.$$
8. So the partial sums of $\sum b_n$ are bounded above, and $\sum b_n$ converges by step 3.