## Comparison Test

*| b*for all

_{n}| | a_{n}|*n > N*. Then the series converges absolutely as well.

If the series
converges to positive infinity, and
is a sequence
of numbers for which
*a _{n} b_{n}*
for all

*n > N*. Then the series also diverges.

for all| b_{n}| | a_{n}|

*n*. then just sum both sides to see what you get formally:

Then:

- If the left sides equals infinity, so must the right side.
- If the right side is finite, the left side also converges.

**Proof:**

The proof, at first glance, seems easy: Suppose that
converges absolutely, and
*| b _{n} | | a _{n} |*
for all

*n*. For simplicity, assume that all terms in both sequences are positive. Let

Then we have thatSand_{N}=T_{N}=

Since the left side is a convergent sequence, it is in particular bounded. Hence, the right side is also a bounded sequence of partial sums. Therefore it converges.T_{N}S_{N}

This proof *wrong*, because it does show that the sequence of
partial sums is *bounded* but it is not necessarily true that
a bounded series also converges - as we know.

However, this proof, slightly modified, does work: Again, assume that all terms in both sequences are positive. Since converges, it satisfies the Cauchy criterion:

if| | <

*m > n > N*. Since

*| b*we then have

_{n}| | a_{n}|if| | | | <

*m > n > N*. Hence satisfies the Cauchy criterion, and therefore converges.

The proof for divergence is similar.