# Prove the convergence of this real sequence and determine its sum

Here is the question:

and here is the 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.

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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

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

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@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