# Series

• Mar 16th 2006, 05:15 PM
Series
"Suppose $\displaystyle \sum a_n, \sum b_n$ are convergent. Let $\displaystyle c_n=\sum_{k=0}^n a_kb_{n-k}$ and let $\displaystyle \sum c_n$ converge. Prove that$\displaystyle (\sum_0^\infty a_n) (\sum_0^\infty b_n) = (\sum_0^\infty c_n).$"

I can't make any headways. Any hints would be appreciated.
• Mar 16th 2006, 05:28 PM
ThePerfectHacker
Quote:

"Suppose $\displaystyle \sum a_n, \sum b_n$ are convergent. Let $\displaystyle c_n=\sum_{k=0}^n a_kb_{n-k}$ and let $\displaystyle \sum c_n$ converge. Prove that$\displaystyle (\sum_0^\infty a_n) (\sum_0^\infty b_n) = (\sum_0^\infty c_n).$"

I can't make any headways. Any hints would be appreciated.

Perhaps you should look at its partial sums.
Let $\displaystyle _aS_n$ represent the nth partial sum of $\displaystyle a_k$ let $\displaystyle _bS_n$ represent the nth partial sum of $\displaystyle b_k$ and $\displaystyle _cS_n$ represent the nth partial sum of $\displaystyle c_k$.
Thus, if we can show that the partial sums of these two series' are equal then they converge to the same number.
Notice that:
$\displaystyle _aS_n=a_0+a_1+...+a_n$
$\displaystyle _bS_b=b_0+b_1+...+b_n$
Then,
$\displaystyle \left(_aS_n\right)\left(_bS_n\right)=(a_0+...+a_n) (b_0+...+b_n)=c_0+c_1+...+c_n$=$\displaystyle _cS_n$
(I might have made a mistake on the last step where I claim it is $\displaystyle _cS_n$ but I do not see anything wrong)
Q.E.D.
• Mar 16th 2006, 06:24 PM