Series

**Treadstone 71** · March 16th 2006, 05:15 PM

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

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

**ThePerfectHacker** · March 16th 2006, 05:28 PM

Originally Posted by Treadstone 71

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

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

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

**Treadstone 71** · March 16th 2006, 06:24 PM

The last step is definately not equal. On one side you have n^2 terms, on the other, you have 1+2+3+...+n terms.

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