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Given that the infinite series $\sum a_n$ converges , we define a new sequence $c_n$ as follows :

Fix $N$ real numbers $b_1,b_2,....b_N$ . Define $$c_k=\begin{cases}b_k ; &\text{for } 1\le k\le N \\ a_k ; & \text{for } k\gt N\end{cases}$$

If $\sum a_n =s$ Then What is $\sum c_n$ $?$

Now $$\sum_{k=1}^{\infty} c_k=\sum_{k=1}^N b_k + \sum_{k=1}^{\infty} a_k \\=\sum_{k=1}^Nb_k +s\\=b+s $$ where $b=\sum_{k=1}^N b_k$.

But the given answer is $$s+b-s_N$$ where $$s_N=\sum_{i=1}^N a_k$$

How do I do that $?$

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    $\begingroup$ $$\sum_{k = 1}^\infty c_k = \sum_{k = 1}^N b_k + \sum_{k = N+1}^\infty a_k$$ $\endgroup$ – Daniel Fischer Nov 17 '15 at 13:16
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You missed the index shift between $(a_k)$ and $(c_k)$ sequence: for $k>N$ you have $c_k=a_k$, so the first $N$ terms of $(a_n)$ are not included in $(c_k)$ but rather replaced with $(b_k)$, hence $s$ is reduced by $s_N$ before advanced by $b$.

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