# Sum of Two Series Proof

• April 27th 2010, 03:22 PM
FailureEqualsLearn
Sum of Two Series Proof
Suppose $\sum_{k=1}^{\infty} (a_k+b_k)$ converges. Then $\sum_{k=1}^{\infty} a_k$ converges <===> $\sum_{k=1}^{\infty} b_k$ converges.

Should I prove this by contradiction using partial sums of the three series? (Headbang)
• April 27th 2010, 03:42 PM
Tikoloshe
What’s confusing is that there is only one case to show here (even though it looks as if not). Without loss of generality, assume one of them converges. Just look at the partial sums to get the final one.

Let $s_n=\sum_{k=1}^n(a_k+b_k)$, $r_n=\sum_{k=1}^n a_k$. Suppose $\lim s_n=s$ and $\lim r_n = r$. A fundamental limit property tells us that $\lim(s_n-r_n)=s-r$. Thus $\sum_{k=1}^nb_k$ converges.
• April 27th 2010, 03:47 PM
Drexel28
Quote:

Originally Posted by FailureEqualsLearn
Suppose $\sum_{k=1}^{\infty} (a_k+b_k)$ converges. Then $\sum_{k=1}^{\infty} a_k$ converges <===> $\sum_{k=1}^{\infty} b_k$ converges.

Should I prove this by contradiction using partial sums of the three series? (Headbang)

Hint:

Spoiler:
The sum of convergent series is convergent $x+(y-x)=y$