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If we let $(a_n)$ and $(b_n)$ be sequences, and suppose there exists $N\in \Bbb N$ such that $a_n=b_n$ for all $n>N$. Suppose $\sum a_n$ converges and $\sum_{n=1}^{\infty}a_n=S$. I am looking to calculate the sum of $\sum_{n=1}^{\infty}b_n$

My solution:

Here is my logic but I am not sure if this is correct thinking or not, looking for some help.

$\sum b_n = S - C$ and $C$ goes to $0$ so $S-(a_1+...+a_N-b_1-b_2-....-b_N)$

2 Answers 2

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Since $\sum_{n = 1}^\infty a_n = S$ and $b_n = a_n$ for any $n > N$ we have $$\sum_{n = N + 1}^\infty b_n = S - \sum_{n = 1}^N a_n$$ Hence $$\sum_{n = 1}^\infty b_n = S - \sum_{n = 1}^N a_n + \sum_{n = 1}^N b_n$$ So what you've done is basically right but $C$ definitely goes NOT to $0$ (it is also a bit unclear how you defined it).

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    so if I took out the c goes to 0 part it would be correct – 2017-01-19
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    @dgreen Yes.... – 2017-01-19
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I don't understand what you mean with the $C$ going to $0$, but the final result is fine.

Another way of going about it would be:

$$\sum_{i=1}^\infty b_i = \sum_{i=1}^N b_i + \sum_{i=N+1}^\infty a_i = \sum_{i=1}^N b_i + \left(\sum_{i=1}^\infty a_i - \sum_{i=1}^{N} a_i\right) = \\ S - \sum_{i=1}^{N} a_i + \sum_{i=1}^N b_i$$

The two sums are finite sums so they exist and $\sum b_i$ also converges.

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    so if I took out the part where c wen to 0 it would be fine? – 2017-01-19
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    @dgreen yes, pretty much. Basically because you start talking about a constant $C$ that you don't define – 2017-01-19