Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is the question:

Question

and here is the solution:

Solution

Can someone please explain it to me - the second line in particular. I don't understand where the limits for the series' come from.

share|cite|improve this question
up vote 1 down vote accepted

First note that

$$\begin{align*} \sum_{n=1}^Kb_n&=\sum_{n=1}^K(b_n-a_n+a_n)\\ &=\sum_{n=1}^K(b_n-a_n)+\sum_{n=1}^Ka_n\;; \end{align*}$$

this is just rewriting $b_1+b_2+\ldots+b_K$ as $$(b_1-a_1+a_1)+(b_2-a_2+a_2)+\ldots+(b_K-a_K+a_K)$$ and then rearranging the terms to get

$$\Big((b_1-a_1)+(b_2-a_2)+\ldots+(b_K-a_K)\Big)+\Big(a_1+a_2+\ldots+a_K\Big)\;.$$

Now assume that $N>K$. Then

$$\begin{align*} \sum_{n=1}^Nb_n&=\sum_{n=1}^Kb_n+\sum_{n=K+1}^Nb_n\\ &=\sum_{n=1}^K(b_n-a_n)+\sum_{n=1}^Ka_n+\sum_{n=K+1}^Nb_n\\ &=\sum_{n=1}^K(b_n-a_n)+\color{red}{\sum_{n=1}^Ka_n+\sum_{n=K+1}^Na_n}&&\text{since }b_n=a_n\text{ for }n>K\\ &=\sum_{n=1}^K(b_n-a_n)+\color{red}{\sum_{n=1}^Na_n}\;. \end{align*}$$

Now take the limit as $N\to\infty$:

$$\lim_{N\to\infty}\sum_{n=1}^Na_n=\sum_{n=1}^\infty a_n\;,$$

so

$$\sum_{n=1}^\infty b_n=\lim_{N\to\infty}\sum_{n=1}^Nb_n=\sum_{n=1}^K(b_n-a_n)+\sum_{n=1}^\infty a_n\;.$$

share|cite|improve this answer
    
@BrianMScott Fantastic answer, that makes things so much clearer. Thank you! – Mathlete Jan 1 '13 at 19:46
    
@Mathlete: You’re welcome! – Brian M. Scott Jan 1 '13 at 19:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.